Method for simulating vertically oriented current in a structure

ABSTRACT

A computer-implemented method of simulating a vertically-oriented current distribution of current flowing through a plurality of layers of one or more three-dimensional conductors embedded in a shielded multi-layered dielectric includes the steps of dividing portions of the circuit into subsections, the portions containing z-directed current into rectangular prisms; independently modeling a current distribution within each subsection, and, specifically, within the rectangular prisms, independently modeling a basis function of linearly changing or uniform current along the z-axis; independently determining the fields resulting from such assumed basis functions; determining a voltage induced by such determined fields, corresponding to a transfer impedance or transfer admittance of the subsection; and calculating a current distribution in one or more conductors according to the transfer impedance or transfer admittance of each subsection and an assumed voltage across each subsection.

CROSS REFERENCE

This application claims the benefit of U.S. Provisional PatentApplication No. 62/848,234, filed on May 15, 2019 and of U.S.Provisional Patent Application No. 62/924,882, filed on Oct. 23, 2019.Both provisional applications are herein incorporated by reference intheir entirety.

BACKGROUND

This disclosure is related to a computer program product and acomputer-implemented method of simulating a current distribution in anarbitrary circuit or other structure embedded in shielded layered media,and, more specifically, simulating the vertically-directed current in ansuch an arbitrary circuit or structure embedded in a shieldedmulti-layered media through the use of tapered vias.

SUMMARY

All examples and features mentioned below can be combined in anytechnically possible way.

According to an aspect, a computer-implemented method of simulating avertically-oriented current distribution of current flowing through aplurality of layers of a three-dimensional structure embedded in ashielded multi-layered dielectric, includes the steps of: dividing thestructure into a plurality of subsections, wherein at least one portionof the structure having vertically-oriented current is divided into atleast one rectangular prism subsection having a first surface disposedperpendicular to a z-axis and a second surface disposed perpendicular tothe z-axis; independently assigning a current in each of thesubsections, wherein z-directed current in the at least one rectangularprism subsection linearly changes from a first value at the firstsurface of the rectangular prism subsection to a second value at thesecond surface of the rectangular prism subsection; wherein the firstvalue is different from the second value; calculating an induced voltagein each of the plurality of subsections resulting from eachindependently modeled current, the induced voltage in each of theplurality of subsections corresponding to a transfer impedance ortransfer admittance; and calculating a current distribution in one ormore conductors according to the transfer impedance or transferadmittance of each subsection and an assumed voltage across eachsubsection.

In an example, the assumed voltage is zero except where a voltage sourceis coupled.

In an example, the assumed voltage is proportional to the current in thesubsection.

In an example, the first value is higher than the second value.

In an example, the first value is lower than the second value.

In an example, z-directed current in a second rectangular prismsubsection is constant throughout.

In an example, both the first predetermined value and secondpredetermined value are nonzero.

In an example, at least one portion of the structure carrying horizontalcurrent is divided into volume rooftop subsections.

In an example, at least one portion of the structure carrying horizontalcurrent is divided into rooftop subsections.

In an example, the current distribution is calculated via a matrixinversion.

According to another aspect, a computer program product stored on anon-transitory computer-readable medium includes a set of non-transitorycomputer-readable instructions for simulating a vertically-orientedcurrent distribution of current flowing through a plurality of layers ofa three-dimensional structure embedded in a shielded multi-layereddielectric, the instructions including the steps of: dividing thestructure into a plurality of subsections, wherein at least one portionof the structure having vertically-oriented current is divided into atleast one rectangular prism subsection having a first surface disposedperpendicular to a z-axis and a second surface disposed perpendicular tothe z-axis; independently assigning a current in each of thesubsections, wherein z-directed current in the at least one rectangularprism subsection linearly changes from a first value at the firstsurface of the rectangular prism subsection to a second value at thesecond surface of the rectangular prism subsection; wherein the firstvalue is different from the second value; calculating an induced voltagein each of the plurality of subsections resulting from eachindependently modeled current, the induced voltage in each of theplurality of subsections corresponding to a transfer impedance ortransfer admittance; and calculating a current distribution in one ormore conductors according to the transfer impedance or transferadmittance of each subsection and an assumed voltage across eachsubsection.

In an example, the assumed voltage is zero except where a voltage sourceis coupled.

In an example, the assumed voltage is proportional to the current in thesubsection.

In an example, the first value is higher than the second value.

In an example, the first value is lower than the second value.

In an example, z-directed current in a second rectangular prismsubsection is constant throughout.

In an example, both the first predetermined value and secondpredetermined value are nonzero.

In an example, at least one portion of the structure carrying horizontalcurrent is divided into volume rooftop subsections.

In an example, at least one portion of the structure carrying horizontalcurrent is divided into rooftop subsections.

In an example, the current distribution is calculated via a matrixinversion.

The details of one or more implementations are set forth in theaccompanying drawings and the description below. Other features,objects, and advantages will be apparent from the description and thedrawings, and from the claims.

FIGURES

In the drawings, like reference characters generally refer to the sameparts throughout the different views. Also, the drawings are notnecessarily to scale, emphasis instead generally being placed uponillustrating the principles of the various aspects.

FIG. 1 shows a perspective view of a circuit disposed in a shieldedmulti-layered media, according an example.

FIG. 2 shows a flowchart of a method for simulating current within acircuit, according to an example.

FIG. 3A shows a perspective view of a rectangular prism volumesubsection, according to an example.

FIG. 3B shows a plurality of plots of vertical current within arectangular prism subsection, according to an example.

FIG. 4A shows a perspective view of a stack of rectangular prismsubsections, according to an example.

FIG. 4B shows a graph of the vertical current within a stack ofrectangular prism subsections, according to an example.

FIG. 5 shows a graph of the vertical current within a stack ofrectangular prism subsections, according to an example.

FIG. 6 shows a perspective view of a circuit disposed in a shieldedmedia, according to an example.

FIG. 7 shows a graph of the modeled circuit over a plurality offrequencies under multiple meshing techniques and numbers ofsubsections, according to an example.

FIG. 8 shows a schematic of the fields surrounding multiple layers,according to an example.

DETAILED DESCRIPTION

The simulation of arbitrary circuits and structures are an importantpart of any electrical design process. Without simulation, suchstructures must be fabricated and individually tested before beingredesigned, fabricated, and tested again. Simulation allows suchcircuits to be redesigned without requiring repeated and costly, timeconsuming fabrication and testing. However, the simulation processitself can be time consuming, requiring hours, even days, to simulate asingle complex circuit with meaningful accuracy (a 50 frequency analysismight require 50 hours or more to complete). This length of simulationtime bogs down design time, slows time-to-market, and, ultimately,represents a costly proposition for electrical design.

Referring now to FIG. 1 there is shown a simple, illustrativethree-dimensional structure 100 in shielded (i.e. surrounded on allsides by a conductor) multi-layered media. The multi-layered media, inthis example, is comprised of four dielectric layers 102, 104, 106, 108,as indicated on the back two sides of the box, although any number oflayers greater than zero may be used. Typically, the layers havedifferent thicknesses, dielectric constants, loss tangents,conductivities, and other constitutive parameters, but they can also bethe same. Furthermore, the layers are planar and have top and bottomsurfaces that are mutually parallel. The sidewalls of the box aretypically fully shielding, i.e., composed of a perfect conductor thatallows no electromagnetic field from the interior to penetrate outside.However, in general, the sidewalls may provide any desired surfaceimpedance boundary condition. For example, a wall composed of a lossyconductor will provide a resistive boundary condition. This can be truefor box sidewalls, as well as the top and bottom covers. For thepurposes of this disclosure, a “shielding” requires being covered ineither a perfect or lossy conductor.

As shown in this example, the structure 100 comprises two thickconductors—bottom conductor 110 and top conductor 112—stacked one on topof the other. The circuit of FIG. 1 is merely provided as an example ofthe type of structure for which the method described in conjunction withFIG. 2 may be used. In alternative examples, the structure 100 canassume any variety of shapes and configurations and have any number oflayers (rather than just two, as shown in FIG. 1). Furthermore, thestructure 100 need not be a conductor at all, but rather, for thepurposes of this disclosure, can be comprised any material differingfrom the embedding media.

A voltage can be applied across the structure 100 to create a currentdistribution within it. In an example, the voltage is a high-frequency(e.g., 10 GHz) time-varying voltage. In the example of FIG. 1 thevoltage can be applied between the left-most face of bottom conductor110 (denoted as face 114) and the immediately adjacent box sidewall.This is known in the art as an ‘infinitesimal gap voltage source’because the gap between which the voltage is applied has zero width. Inthis example, the box sidewall can be viewed as ‘ground.’ The voltagecould also be applied between the rear-most face of top conductor 112(denoted as face 116) and the immediately adjacent (back) sidewall.(Although the voltage can be applied across any two points of anyportion of the structure). In both cases the voltage exists between thecircuit and ground. This is how nearly all circuits are excited for EManalysis. In an alternative example, however, the voltage can be appliedbetween any two points in a circuit (such as between face 114 and face116).

As a result of this applied voltage, a current distribution, whichincludes both horizontal current and vertical current, forms withinstructure 100. The portion of the structure 100 in which only horizontalcurrent flows can be accurately modelled, for example, using Method ofMoments, realized with infinitely-thin rooftop subsections, volumerooftop subsections, or other known subsections.

The junction region where the bottom conductor 110 and top conductor 112overlap (i.e., more generally, where the layers of the structureoverlap), however, requires vertical current, which is typically modeledusing stacks of “uniform vias” (blocks of constant vertical current).Stated differently, the current in a given uniform via is verticallydirected and is the same magnitude and phase everywhere. Stacks ofuniform vias are thus used in a Method of Moments analysis to simulate apiecewise-constant representation of the vertical current, which is thenused to solve for the current distribution of vertical current withinthe circuit. But uniform vias are poor representations of the actualvertical current distribution, which in reality gradually and smoothlychanges in both magnitude and phase as it flows along a long via. Thus,in order to accurately model vertical current, a large number of veryshort uniform via blocks are required to generate a piecewise-constantrepresentation with sufficient resolution to achieve the desiredsimulation accuracy.

For complex circuits, the large number of vias required to accuratelymodel vertical current is computationally prohibitive. Matrix solve isthe main numerical bottleneck in this analysis. Matrix solve is an N³operation where N is the number of subsections, meaning that eachmodeled subsection exponentially increases the computational complexityand time required to simulate the vertical current distribution. Thecomputational complexity to solve structures such as a large array ofthrough-silicon-vias (TSVs), for example, is prohibitive. There exists,then, a need in the art for a computer-implemented method that candrastically reduce the number of subsections, and thus, time, requiredto model the z-directed current distribution within a three-dimensionalstructure disposed within a shielded multi-layered media. So-called‘iterative matrix solves’ can result in faster solve times, but, inspite of decades of research, are still lacking in robustness,consistent ability to reach convergence, and reliable accuracyespecially for shielded structures such as are considered here.

Accordingly, various examples described in this disclosure are directedtoward an efficient computer-implemented method for modeling thez-directed current distribution within a three-dimensional structuredisposed within a shielded multi-layered media. More specifically,various examples describe using subsections of linearly-changingvertical current. Using such subsections, rather than only uniformsubsections, the number of subsections required to model verticalcurrent may be reduced by a factor of approximately 6.6, reducing thenumerical complexity by approximately 280 times, thus allowing acomputer to solve for the vertical current distribution approximately280 times faster. Stated differently, an analysis which would requireone hour using only uniform vias may be reduced to 13 seconds, and ananalysis that would require 50 hours can be reduced to under 11 minutes,while retaining the same degree of accuracy.

For the purposes of this disclosure, the x-y axes are oriented parallelto the surfaces of the dielectric layers and the z-axis is orientedperpendicular to the surfaces of the dielectric layers. Z-directedcurrent (alternatively referred to in this disclosure as “verticalcurrent”), then, is current that flows in a direction perpendicular tothe surfaces of the dielectric layers. Z-directed current can cross thetop and bottom surfaces of the dielectric layers, flowing between layersof the dielectric. A structure that connects multiple layers of thestructure disposed in different dielectric layers is known as a via.Thus, any via that carries current will be carrying current that has, atleast, a z-directed component. It should be understood that thedesignations of an “x-axis,” a “y-axis,” and a “z-axis” are somewhatarbitrary and can be interchanged without affecting the underlyingconcept—the axes are so defined to represent a useful way for describingthe orientation of the rectangular prisms and the direction of currentwithin a circuit. For example, the axes could be reoriented such thatcurrent flowing perpendicular to the surface of the dielectric layersflowed parallel to the x-axis, without changing the underlying conceptdescribed herein. In fact the choice of coordinate system to describethis problem is completely arbitrary. Any choice may be made and theunderlying physics remains unchanged.

FIG. 2 shows a flowchart of such a computer-implemented method 200 forsimulating a vertically-oriented current distribution of current flowingthrough a plurality of layers of a three dimensional structure (e.g.,structure 100) embedded in a multi-layered dielectric, which iscomputationally efficient and, thus, faster to implement than previousmethods. The computer may be a general-purpose computer comprising aprocessor and a non-transitory storage medium containing instructionsthat, when executed by the processor, carry out the steps described inmethod 200; however, it should be understood that any computing device,suitable for carrying out the steps described in connection with FIG. 2can be used.

At step 202, the arbitrary three-dimensional circuit is divided intomany subsections that are typically small (the range of subsection sizescan extend from a few millimeters down to a fraction of a micron) withrespect to the wavelength (e.g., 1/10^(th) of a wavelength or 1/100^(th)of a wavelength) of the voltage applied to the planar circuit.Typically, the subsection size is adjusted to be small at the highestfrequency (shortest wavelength) and then the same size subsection isused at all frequencies. In practice, the smallest subsection size isactually set by the largest size that allows subsectioning of the finestportions of the circuit. This often turns out to also be small withrespect to wavelength. For Method of Moments, only the structure (again,everything in the structure that is different from the embeddingmulti-layered media) is meshed; the embedding dielectric layers (as usedin this disclosure, “dielectric” includes lossless and lossy dielectric,as well as semiconductors like silicon) and the shielding box are notmeshed. As is described below, the subsections can be rooftopsubsections, volume rooftop subsections, rectangular prism subsections,or any other suitable subsection.

As part of this step, any via (i.e., portion of the circuit throughwhich current can flow between portions of the structure disposed inadjacent dielectric layers) or any portion of the structure whichotherwise carries z-directed current during use, can be subdivided intorectangular prisms (or at least one rectangular prism). Consider, forexample, the region of transition in FIG. 1, where the lines overlap andtouch. This region can be divided into two rectangular prismsubsections, designated in FIG. 1 as subsections 118 and 120. The bottomsubsection 118 envelops the right end of the lower line. The topsubsection 120 envelops the near end of the upper line. It is in thesetwo regions that vertical (z-directed) current flows.

A rectangular prism is defined as a three-dimensional object which hassix faces that are rectangles. One such rectangular prism subsection isshown in FIG. 3A. As shown, the top and bottom surfaces of therectangular prism are arranged parallel to the x-y plane, and the sidesurfaces (i.e., those surfaces other than the top and bottom surfaces)are arranged either perpendicular to the x-axis (i.e., parallel to they-z plane) or perpendicular to the y-axis (i.e., parallel to x-z plane).Which surface of the surfaces parallel to the x-y plane is identified asthe “bottom” surface or the “top” surface is largely for the purpose ofconvenience in explanation, and the names may be interchanged (as longas they are done so consistently throughout) without affecting theunderlying concept. The sides of the rectangular prism are thus arrangedparallel to the z-axis.

In the example of FIG. 3A, a single subsection is shown to cover an areaof 24 x by 24 y and extends over a thickness of h, corresponding to thethickness of the layer in which the subsection is disposed. Thesubsection, as shown, thus extends across the width of the via andacross the height of the dielectric layer. It should, however beunderstood that, in various alternative examples, the subsection canextend across only a portion of the dielectric, and thus h may be someheight less than the thickness of a dielectric layer. Furthermore, thesubsection can have a thickness h greater than a particular dielectriclayer and thus can extend across multiple layers. In this disclosure, arectangular prism subsection within a via is alternatively referred toas a “via block,” or simply a “block.” In addition, not only can a viablock extend over multiple dielectric layers, it can also extend overmultiples of 2Δx by 2Δy.

Generally, the values of Δx, Δy, and h are selected such that thesmallest possible via extends over just 2Δx by 2Δy and over thethickness of the layer and is called a ‘basis function’ (this will bedescribed further, below). It should be understood that the measurements2Δx and 2Δy are selected for purposes of convenience. Conceivably, asubsection could extend over some distance besides 2Δx and 2Δy. In suchan instance, the values of Δx and Δy are selected to extend over thesame distance that the 2Δx and 2Δy values would have otherwise covered.For example, a subsecution could extend over just one delta Δx and Δy,in which case, the values of Δx and Δy would be double the values of Δxand Δy when 2Δx and 2Δy is used for a single subsection. (Likewise, ifthe subsection extends over 0.5Δx and 0.5Δy, the values of Δx and Δywould be quadruple the values of Δx and Δy when 2Δx and 2Δy is used fora single subsection.) In other words, the physical meshing would be thesame, the only difference is the “yardstick” being used to measure it.Similarly, the thickness of a subsection could extend over a fraction ofa layer, in which case the layer in which the subsection is disposed isconceptually split into multiple layers with one layer thickness equalto a subsection height. Again, the situations would be physicallyequivalent—the only thing changing is the size of the layer thickness“yardstick.” Multiple basis functions can be added together side-by-sideand one-on-top-of-the-other and get a large via subsection, or use asingle basis function for the smallest possible via subsection. In orderto maximize the use of FFT in the numerical calculations that follow inthe steps below, it is preferable that all via basis functions use thesame values for Δx and Δy. However, different vias in the same structurecan use different values for Δx and Δy, but this will limit or eliminateuse of FFT and thus slow the calculation time.

At step 204, each subsection is independently assigned a predeterminedcurrent distribution typically having both a magnitude and a phase.(“Independently,” for the purposes of this disclosure, means“separately.” Thus, the analysis that follows in steps 206 and 208results from the current distribution assigned to a single subsectionand is similarly separately performed for each remaining subsection. Theanalysis, however, of each separate subsection can occur simulataneouslyor concurrently). The current distribution on the subsection assumes aspecific form that is a sum of one or more fundamental currentdistributions known as a ‘basis functions’ or ‘expansion functions’ andis typically assumed to have a peak current density of unit magnitude(e.g., 1 A/m² for volume current basis functions), although othermagnitudes can be used. The basis function assigns the shape of thecurrent distribution on each subsection. In addition to magnitude, aphase is typically assigned to each subsection, thus current can have aphase shift across a conductor, moving from one basis function to thenext. For example, one basis function might be at phase 0 degrees, thenext at 10 degrees, etc.

The remaining steps described below are used to numerically evaluate themagnitude and/or phase of the current for each basis function. Themagnitude and phase is, in certain examples, selected so that the totalinduced voltage on all subsections (except where an input voltage isapplied) goes to zero. The total current on the subsection is then thebasis function (which gives it the shape of the current distribution)multiplied by the determined magnitude and/or phase.

Horizontal current can be simulated using a mixture of any known basisfunctions for each type of subsection. The smallest possible subsectionconsists of a single basis function. For example, the basis functionsfor rooftop subsections are known and one or more may be joined togetherin any way desired to construct each (typically larger) rooftopsubsection. Likewise, for volume rooftop subsections the basis functionsfor volume rooftop subsections can be used.

The basis function within at least one rectangular prism subsection thatincludes vertical current can be assigned a tapered via basis function.If the rectangular prism of FIG. 3A exists at the bottom of a via (e.g.,bottom subsection 118), the vertical current on the bottom surface ofthe rectangular prism must be zero because it is an open circuit. Thus,only horizontal current can flow there. At the top surface of the bottomblock (where the bottom and top via blocks touch), the vertical currentwithin this subsection will be at a maximum. This is where all thepreviously horizontal current from the subsection (over the entirethickness of the bottom conductor 110) is now flowing vertically toconnect with the near end of the top line (at the top of top conductor112). The tapered via can be viewed as picking up and gathering all ofthe horizontal current and converting it to vertical current. This iswhy it is linearly tapered. For example ½ way up the tapered via, thetapered via will have gathered ½ of the total incoming horizontalcurrent. Thus, in order to model the vertical current in the bottomblock, a via subsection with zero current at one surface in the x-yplane (in this example, the bottom surface) and maximum current at theopposite surface in the x-y plane (in this example, the top surface),i.e., a tapered via, can be used. This example, in which verticalcurrent is at a minimum at the bottom surface and at a maximum at thetop surface, may also be referred to as an ‘up via’ or an ‘up-taperedvia’ or a ‘tapered up-via’ as it is often used to transition currentupwards from a horizontal volume rooftop.

Generally speaking, a subsection can be any size that contains aninteger number of basis functions. In this manner, basis functions canbe conceived of as building blocks. In practice, all the basis functionsof a given type (like a tapered via, or a volume rooftop) are all thesame size. In theory, however, any size basis functions can be mixed(i.e., using different sizes all in the same analysis; using basisfunctions of differing size, however, limits the use of the FFT and thusslows the calculation.

In FIG. 3B, plot 302, shows the magnitude of the vertical volume currentwithin a via block when a tapered volume up via subsection (i.e., basisfunction) occupies that block. The current is the same everywhere in anyhorizontal cross-section, and it tapers linearly from zero at the bottomto a maximum at the top, or, assuming the base of the block is at z=0,

$\begin{matrix}{J_{Z} = \frac{Z}{h}} & 1\end{matrix}$

Note that the ‘taper’ refers only to the vertical current densityvariation within the via. The physical shape of a tapered via is arectangular prism, as described above in connection with FIG. 3A. Thus,at least one of the rectangular prism subsections in the via orz-directed current portion of the circuit (as determined in step 202) isassumed to contain current changing linearly (in magnitude and/or phase)in the z-direction. Thus, current density in a tapered via is linearlychanging in magnitude and/or phase in the z-direction.

The nature of the assigned z-directed current distribution—whether itlinearly increases from the bottom surface up to the top surface,linearly increases from the top surface down to the bottom surface, orremains uniform from the bottom surface to the top surface—is selectedbased on the respective location of the rectangular prism subsectionwithin the circuit.

For example, consider now the vertical current in the top block (e.g.,top subsection 120), at the near end of the top line. The top surface ofthe top block must have zero vertical current. All current flow hastransitioned to horizontal current flow in the volume rooftops used tomodel the top line. The bottom surface of the top block, where it meetsthe top surface of the bottom block, must have maximum vertical currentas all the previously horizontal current is flowing vertically acrossthis surface. Thus, a tapered via that has zero current at the top andmaximum current at the bottom in order to model the vertical componentof current in this region is required. This is referred to in thisdisclosure as a ‘down via’, a ‘down-tapered via’, or a ‘tapereddown-via’ as it is often used to transition current downwards.

The current distribution in a via block disposed at the center of a viastack, where the current at the top surface is typically equal tocurrent at the bottom surface, can be modeled as a “uniform via.” Auniform via has constant vertical current throughout, as shown in FIG.3B plot 304. For example, a via block will have a uniform currentdistribution if the via block is the center block within an odd numberof stacked via blocks (such as is shown, for example, FIG. 4A) with thetop and bottom ends of the stack open circuited, as in a resonator(often used in filters) or a dipole (an antenna).

FIG. 3B plot 306 shows the current distribution for a tapered via and auniform via occupying exactly the same volume, i.e., combined bysuperposition. The total current in the via volume is the (weighted) sumof the two currents, which allows modeling of a current that linearlytapers from one value at the bottom of the via (as determined by theuniform via as the up-tapered via has zero current here) to anothervalue at the top of the via (as determined by the uniform via plus themaximum current present in the up-tapered via). This is required, forexample, when a length of the structure extends over three or morelayers. The via at the bottom end will require only an up-tapered via(which has zero current at the bottom end). Superposition allowsmodelling current with any slope at any level. The linearly tapered viacurrent (plot 302), when combined in the same volume with a uniformcurrent via (plot 304), allows representation of via current that varieslinearly from one value at the bottom of the via to another value at thetop of the via (plot 306). All interior via blocks typically requireboth a tapered and a uniform via, adding together as shown in plot 306,in order to represent the non-zero current at either end of any giveninterior via block. When both the tapered via and the uniform via arecollocated in the same via block and also treated as separatesubsections, application of Method of Moments results in the calculationof how much current each subsection should have to best represent theactual current distribution in the circuit being analyzed. The top layerrequires zero current at the top end, which can be represented by a downvia or by an up-tapered via being subtracted from a uniform via, andthus canceling to zero at the top.

In general, a piecewise-linear representation of the current flowing ina long via is realized by dividing the via into multiple uniform plustapered via subsections (i.e., a stack of via blocks, described hereinas a “via stack”). When the current is modeled flowing from a thickconductor on one level to a thick conductor on another level, as in FIG.1, this allows the horizontal current from a volume rooftop totransition to vertical via current and, on the next layer, to transitionback to horizontal current. A transition of current from a volumerooftop subsection on one layer up to the next layer cannot beaccomplished with a uniform via at all. A tapered via is required.Uniform vias have long been in use to transition current from oneinfinitely thin rooftop to another infinitely thin rooftop on anadjacent layer. The tapered via also models current changing smoothlyalong the length of a long via due to high frequency electromagneticeffects. This is in sharp contrast to uniform vias which allow only apiecewise-constant model of what is in reality a smoothly changingcurrent along the length of the via.

An example via stack, yielding such a piecewise linear representation ofthe current on the via stack, is shown in FIG. 4A. The via stack of FIG.4 includes five via blocks 400, 402, 404, 406, and 408. (Although eachvia block shown in FIG. 4 is of the same height, it should be understoodthat, in alternative examples, the via blocks can be of various heightsand need not be the same.) As shown in FIG. 4B, the bottom end of thestack contains only an up via (410). The top end contains only a downvia (418). All other blocks contain both an up via (or, alternately adown via) and a uniform via (412, 414, 416). A lengthy (compared towavelength) via can be accurately modelled by such a stack of viablocks.

Note that block 406 in the via stack of FIG. 4 appears to be modelled asa superposition of a uniform via and a down via (416). However, a downvia is not actually required. If a block is occupied by both an up viaand a uniform via, Method of Moments assigns a negative weight to the upvia when a down via is required. This subtracts the up via current fromthe uniform via current creating the effect of a down via.Fundamentally, a down via is an up via subtracted from a uniform via.Thus, of the three types of vias, down, uniform, and up, only two shouldbe used in any given volume block as the third is a linear combinationof the other two. Use of all three via types results in a singularmoment matrix. Matrix inversion, and thus the entire analysis, failswhen this happens. Here, only the uniform via and up via are used. If adown via is needed it is formed by subtracting an up via from a uniformvia. Alternatively, instead of an up via, only a down via and a uniformvia may be used. Likewise, in a different example, only a down via andan up via may be used, where the uniform via is simulated as asuperposition of an up via and a down via.

There are many via basis functions that can be used to obtain a taperedvia by superposition. In general any selection of two vias with lineartaper plus offset, as in 306 in FIG. 3B (even including zero offset, asin 302, or zero slope, as in 304) may be used as long as one is not‘linearly dependent’ on the other, i.e., one cannot be written as asimple multiple of the other. For example, two linear tapered up-vias,both starting at zero but with different tapers are linearly dependentand not suitable. However, an up-tapered via starting at zero and asecond up tapered via starting at 0.5, are linearly independentregardless of the slope of their respective tapers. In other words, theup-tapered, the down-tapered, and the uniform via may all be written asa weighted sum of those two vias. Any set of via basis functions thatinclude a linear taper and may be linearly combined, i.e., byapplication of superposition to obtain the tapered vias described hereis deemed to be equivalent to the tapered via described in thisdisclosure.

Steps 206 and 208 are analytical developments, i.e., performed in termsof general equations. These equations are developed once and thenexecuted by a computer processor following the steps of a computerprogram. The equations are written so that they may be applied to anyspecific kind of basis function in any location within the shieldedlayered media. After implementation in a computer program, the computerprogram then applies the equations to all the corresponding types ofsubsections (which are formed from one or more basis functions) requiredfor analysis of a specific circuit.

At step 206 the fields surrounding each subsection due to the assignedbasis function current, including the effect of all layers of dielectricand the effect of the surrounding conducting walls, but excluding allother circuit metal, are independently determined. For the purposes ofthis disclosure, individual basis functions are determined for allsubsections. This analysis is easily extended to subsections composed ofmultiple basis functions by invoking superposition. The subsection withthe assumed current distribution is referred to as the ‘source’subsection and can be viewed conceptually as a tiny transmit antennawithin the shielded layered media.

Derivation of the reaction integrals used by Method of Moments starts byderivation of the full dyadic Green's function in terms of waveguidemodes, where the waveguide modes are defined for each dielectric layerby the shielding box sidewalls. Next, the Green's function is multipliedby the tapered via current density and integrated over the volume of thetapered via. This determines the fields that surround the tapered via.

In general, this analytical portion of the analysis is based onrepresenting the fields surrounding a tapered via as a weighted sum ofall possible TE (transverse electric, i.e., no z-directed electricfield) and TM (transverse magnetic, i.e., no z-directed magnetic field)rectangular waveguide modes, with the conducting sidewalls of the box(FIG. 1) forming a rectangular waveguide that propagates waves in thez-direction. While the sidewalls are typically perfectly conducting,sidewalls with any arbitrary surface impedance can be implemented byappropriate modification of the waveguide modes being used. Theserectangular waveguide modes, together, form a complete orthogonal basisfor representing all possible fields in the shielded-layered media. Whenthe source subsection has only z-directed current, the TE waveguide modeamplitudes are all zero. Z-directed current generates only TM waveguidemodes.

The basic variables for this analysis are defined using standardrectangular waveguide terminology and are detailed in eq. (1)-(11) inAppendix 1, below. The rectangular waveguide is formed by the conductingsidewalls of the shielding box with the dielectric layers viewed asforming a (vertical) cascade of homogenous rectangular waveguides. Also,as is standard practice for most RF (radio frequency) and microwavework, fields are assumed to have a time-harmonic dependence of exp(jωt),where ω is the radian frequency of the time-harmonic wave. When thisterm is substituted into the governing differential and integralequations (also known as Maxwell's Equations) it cancels out of allexpressions leaving only the complex (i.e., real and imaginary)amplitudes of the various time-harmonic waves. Thus, in all workpresented here, only the complex amplitude of all waves at eachfrequency need be considered.

The Fourier series amplitudes for both a square and a triangular pulseare also required. This is a result of the integral of the basisfunctions (which are formed from rectangular and triangular pulses, FIG.3B plot 302 shows a rectangular pulse and 304 shows the first half of atriangle pulse) multiplied by the sines and cosines of the rectangularwaveguide modes. The required Fourier coefficients follow.

$\begin{matrix}{{F_{R}\left( {\Delta x} \right)} = \begin{matrix}{{\frac{2}{k_{x}}\sin\left( {k_{x}\Delta x} \right)},} & {k_{x} \neq 0} \\{{\Delta\; x},} & {k_{x} = 0}\end{matrix}} & 2 \\{{F_{T}\left( {\Delta x} \right)} = \begin{matrix}{{\frac{1}{\Delta xk_{x}^{2}}\left( {1 - {\cos\left( {k_{x}\Delta x} \right)}} \right)},} & {k_{x} \neq 0} \\{{2\Delta\; x},} & {k_{x} = 0}\end{matrix}} & 3\end{matrix}$

The subscript R refers to the Fourier coefficients for a rectangularpulse, (FIG. 3B, plot 304). The subscript T refers the Fouriercoefficients for a triangle pulse, a ramp-up (FIG. 3B plot 302)immediately followed by a ramp down, in this case, in the x-direction.The horizontal rooftop function current distribution is a triangle pulse(including both ramp-up followed by a ramp-down) in the direction of thecurrent flow and a rectangle pulse variation in the directionperpendicular to the current flow.

The above Fourier coefficients are for a rectangle or triangle pulsevariation in the x-direction. For a pulse in the y-direction, change Δxto Δy and k_(x) to k_(y). A rooftop with x-directed current will have atriangle pulse variation in the x-direction and a rectangle pulsevariation in the y-direction with corresponding Fourier coefficients ofF_(T)(Δx) and F_(R)(Δy).

The tapered volume via results use the Fourier coefficients in thefollowing expressions:

F _(VIA)(x,y)=F _(R)(Δx)F _(R)(Δy)N ₃ g ₃(x,y)  4

F _(RFX)(X,y)=F _(T)(Δx)F _(R)(ΔY)N ₂ g ₁(x,y)  5

F _(RFY)(x,y)=F _(R)(Δx)F _(T)(Δy)N ₁ g ₂(x,y)  6

Eq. (4) is the Fourier coefficients for a VIA (with the samecoefficients used for both tapered and uniform vias), which uses arectangular pulse for the current distribution in both x-andy-directions. Eq. (5) is the Fourier coefficients for a rooftop basisfunction with x-directed current, RFX, which uses a triangle pulse (rampup followed by ramp down) in the x-direction and a rectangle pulse inthe y-direction. Both surface (i.e., infinitely thin) and volumerooftops use the same coefficients. Eq. (6) is the Fourier coefficientsfor a rooftop with y-directed current, RFY., in which N₁ is substitutedfor N₂ and g₂ is substituted for g₁. The N coefficients are normalizingconstants and the g functions contain the x and y varying sines andcosines of the waveguide modes. These are both detailed in the first fewequations of Appendix 1. Note that the waveguide mode numbers in theseequations and in certain other equations below are not explicit, forclarity. When the standard waveguide constants are substituted in forthese and other expressions, for example, for k_(x) and k_(y), then themode numbers, m and n, become explicit.

The results also include the following term

D=(1−r _(IT) r _(iB))sin(k _(iz) h)+(r _(iT) +r _(iB))cos(k _(iz) h)  7

as a denominator. The index i represents the waveguide mode numbers andis used to sum over all TE and TM m and n modes. However, for via sourcesubsections, which have only z-directed current, all TE modes are zero.Thus our sums, in this disclosure, are all over TM modes only.

The r_(iT) and r_(iB) are the normalized surface impedances present onthe top (T) and bottom (B) surfaces of the dielectric layer containingthe volume via subsection. If perfectly conducting ground planes arepresent on both surfaces, then r_(iT) and r_(iB) are both zero. If thereare additional layers of dielectric between either cover and thevia-containing layer, the surface impedances of the covers aretransformed by standard rectangular waveguide transmission line theorythrough the intervening dielectric layers to the source layer's top andbottom surfaces. Details and a full derivation are provided in Appendix1 and values are determined recursively using eq. (23)-(26) and(31)-(34).

At step 208, equations for the voltage induced in the location of allother subsections are independently determined based on the fieldequations that were determined in step 206. The other subsections (i.e.,those subsections not viewed as the source subsection in this step) canbe viewed as tiny receiving antennas and are called ‘field’ subsections.To calculate the voltage induced in each field subsection, the electricfield due to the source subsection is integrated at the location of andover the volume of the field subsection. This is known as a ‘reactionintegral’ and can be viewed as the electromagnetic coupling between twosubsections. It is convenient, although not strictly necessary, to firstmultiply the electric field by a weighting function prior tointegration. For example, if the weighting function is a delta function,such that the electric field is effectively sampled at a specified pointand without integration, the ‘point matching’ variety of Method ofMoments results. For this work, Method of Moments is illustrated byselecting as a weighting function the basis function that willeventually be used to represent current on the field subsection. Thisresults in the Galerkin technique′ or Galerkin′ implementation of theMethod of Moments. In addition, the source subsection is itself alsoviewed as a field subsection and the ‘self-coupling’, i.e., the voltageinduced in the source subsection by the current on the sourcesubsection, is determined. These equations, which have been determinedfor the electromagnetic coupling between any possible pair of specifictypes of basis functions with arbitrary locations within the shieldedlayered media are implemented in a computer program.

The result is an equation with an exact analytical infinite summationfor the integrated voltage on each field subsection that exactlyincludes the effect of the dielectric layers and the box. The summationsare typically performed numerically using the FFT (Fast FourierTransform) which is easily organized so that all summations areperformed to nearly full numerical precision. Thus, the summationresults are described as ‘numerically exact.’ (As will be understood bya person of ordinary skill in the art, the term “numerically exact”differs from a truly “exact” calculation. An “exact” computation wouldrequire the computation of an infinite number of items. Thiscalculation, however, can be truncated at some point after the terms ofthe infinite sum become so small that the differences no longer matter.)The exact equations for these summations are listed below. Inalternative examples, the summations could also be done directly, i.e.,without the use of the FFT algorithm, by calculating each term andadding them up. This is generally less computationally efficient, butcould be required if different values of Δx and/or Δy are used indifferent places in the circuit.

This summation is repeated for every subsection, letting each one takeits turn as a ‘tiny transmitting antenna,’ source subsectionindependently. This calculation includes the voltage induced in thesource subsection by itself. Once complete, all this information aboutcoupling between all possible pairs of subsections can be stored in amatrix. (A matrix represents a convenient way to solve a system ofequations, and, in alternative examples, the same system of equationscan be solved without the use of a matrix.) For example, if 1.0 amperesplaced on subsection number 18 induces, or couples, 2.318654876 volts insubsection number 32, then the (32, 18) element of the matrix is2.318654876. This matrix is called the ‘moment matrix’. So the (matrix)equation is now: (Voltage on Each Subsection)=(Big MomentMatrix)*(Current on Each Subsection). Notably, this is just a matrixversion of Ohm's law, with the induced voltage representing a transferimpedance from the current on each subsection. (In an alternateembodiment, the matrix may be inverted to represent transferadmittance.) The advantage of using a Galerkin technique is that thematrix is symmetric. Thus the (18, 32) element of the matrix takes thesame value. Symmetric matrices can be stored with half the memory andrequire half the time to fill as compared to a full matrix.

Stated differently, the result of step 208 is equations that, given ameshing, or subsectioning, of a specific circuit, allow a computerprogram to calculate numerical values for an N×N matrix equation: V=Z*I,where V and I are Nx1 column vectors and Z is an N×N matrix, which iscalled an impedance matrix. The moment matrix is Z, which at this stageof the analysis is an impedance matrix. The first row of that matrixequation may be written as V₁=Z₁₁*I₁+Z₁₂*I₂+Z₁₃*I₃+ . . . , where V₁ isthe total voltage induced in subsection 1, which is the sum of voltagesinduced in subsection 1 by the current on all subsections. The meaningof this equation may be explained as follows: The current on eachsubsection is the basis function(s) of that subsection (determined instep 204) multiplied by the overall current amplitude as set by thevector I. Thus the actual current on subsection 1 is the current assumedon the basis function(s) that compose subsection 1 multiplied by thefirst element of the I vector, I₁. The voltage on subsection 1 due tocurrent on subsection 1 is Z₁₁ multiplied by I₁. This is the voltagethat the current on subsection 1 induces on itself, or the‘self-coupling’. In an analogous manner, the voltage on subsection 1 dueto current on subsection 2 is Z₁₂ multiplied by 1₂. In a similar manner,Z₁₃ multiplied by 13, the current on subsection 3, is the voltageinduced across subsection 1 by the current on subsection 3. This isrepeated for the current on each of the remaining subsections. Theentire voltage summation process is then repeated for each of theremaining subsections V₂, V₃, V₄, etc. However, this process works onlyif the current on each subsection as specified by the I vector is known.The values of the I vector are not known in advance: the I vector isactually the unknown being solved for. Fortunately, the values that thevoltage vector must be are known: the entire voltage vector must be zero(no voltage is allowed across a perfect conductor) except for thesubsection upon which we apply a known input voltage. Therefore, thismatrix equation is solved for the unknown current vector in terms of theknown voltage vector. This can use any of the many well-known matrixsolve techniques, for example, ‘LU Decomposition’ or ‘GaussianElimination’, etc. This takes place in step 210, as described below. Useof the so-called ‘iterative matrix solves’ at this point and for thisspecific type of problem are not yet sufficiently robust or accurate forall possible circuits.

The exact analytical equations for the Method of Moments reactionintegrals, here denoted by S, even though they are used to fill theMethod of Moments Z matrix, for tapered volume vias are listed below.Again, these are the voltages induced in a field subsection by currentin a source subsection. These equations are evaluated for pairs of thespecified basis functions at the specified locations within the layerwith results used to fill the Method of Moments Z matrix and form thesolution to the central problem in formulating a method of momentsanalysis. Derivations are fully detailed in Appendix 1. It should beunderstood that the equations presented in this disclosure assume aspecific form and can be presented in many different forms equivalent tothe forms used herein.

The reaction integral for coupling between a tapered via (TV) centeredat (x₀,y₀) that extends vertically over the entire layer thickness, h,and an infinitely thin (horizontal) rooftop with x-directed current(RFX) centered at (x₁,y₁,z) where z is the level at which the infinitelythin rooftop is placed and is within or on the surface of the same layerfollows (summation is over all TM modes only) (eq. 194 and 191 inAppendix 1).

$\begin{matrix}{\mspace{79mu}{S_{TVtoRFX} = {\sum{\frac{{F_{VIA}\left( {x_{0},y_{o}} \right)}{F_{RFX}\left( {x_{1},y_{1}} \right)}}{Y_{i}k_{iz}h}\left( {1 - \frac{z_{3}(z)}{D}} \right)}}}\ } & 8 \\{{Z_{3}(z)} = {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right.}}} & 9\end{matrix}$

For coupling between a tapered via centered at (x₀,y₀) and infinitelythin rooftops with y-directed current (RFY) centered at (x₁,y₁),substitute F_(RFY) for F_(RFX) in the above expression.

The reaction integral for coupling between a tapered volume via centeredat (x₀,y₀) and a volume rooftop with x-directed current (VRFX) centeredat (x₁,y₁), both extending over the entire thickness, h, of the samelayer follows (eq. 203 in Appendix 1).

$\begin{matrix}{\mspace{79mu}{S_{TVtoVRFX} = {\sum{\frac{{F_{VIA}\left( {x_{0},y_{o}} \right)}{F_{RFX}\left( {x_{1},y_{1}} \right)}}{Y_{i}k_{iz}^{2}h}\left( {{k_{iz}h} - \frac{Z_{3V}(h)}{D}} \right)}}}\ } & 10 \\{{Z_{3V}(h)} = {{\left( {2 - {r_{iT}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} + {\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right){\sin\left( {k_{iz}h} \right)}}}} & 11\end{matrix}$

For coupling between a tapered via centered at (x₀,y₀) and a volumerooftop with y-directed current (VRFY) centered at (x₁,y₁), substituteF_(RFY) for F_(RFX) in the above expression.

As with all reaction integral results in this disclosure, the summationover the index i is taken over all TM modes only. For clarity, the modalsummation index i, which sums over all m, n waveguide modes, is notalways explicitly given, but can be easily determined from the standardwaveguide mode definitions detailed in eq. (1)-(11) in Appendix 1,below.

The reaction integral for coupling between a tapered volume via centeredat (x₀,y₀) and a uniform volume via (UV) centered at (x₁,y₁), bothextending over the entire thickness, h, of the same layer, follows (eq.232 in Appendix 1).

$\begin{matrix}{S_{TVtoUV} = {\sum{\frac{{F_{VIA}\left( {x_{0},y_{o}} \right)}{F_{VIA}\left( {x_{1},y_{1}} \right)}}{y_{i}}\left( {\frac{Z_{4U}(h)}{D} + {\frac{k_{iz}h}{2}\frac{j\omega\mu k_{iz}^{2}y_{i}}{k_{ic}^{2}}}} \right)}}} & 12 \\{\mspace{79mu}{{Z_{4U}(h)} = {{\left( {\frac{r_{iT} - r_{iB}}{k_{iz}h} + {r_{iT}r_{iB}}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} - {r_{iT}{\sin\left( {k_{iz}h} \right)}}}}} & 13\end{matrix}$

The reaction integral for coupling between a tapered volume via (TV)centered at (x₀,y₀) and another tapered volume via centered at (x₁,y₁),both extending over the entire thickness, h, of the same layer follows(eq. 244 and 240 in Appendix 1).

$\begin{matrix}{S_{TVtoTV} = {\sum{\frac{{F_{VIA}\left( {x_{0},y_{o}} \right)}{F_{VIA}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( {\frac{Z_{6}(h)}{D} + {\frac{k_{iz}h}{3}\frac{j\;{\omega\mu}\; k_{iz}^{2}Y_{i}}{k_{ic}^{2}}}} \right)}}} & 14 \\{{Z_{6}(h)} = {\frac{1}{k_{iz}^{2}h^{2}}\left\lbrack {{\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right)k_{iz}h} + {\left\{ {{\left( {{r_{iT}r_{iB}} + 1} \right)k_{iz}h} - {r_{iT}k_{iz}^{2}h^{2}} - r_{iT} - r_{iB}} \right\}{\sin\left( {k_{iz}h} \right)}} + {\left\{ {2 - {\left( {r_{iT} - r_{iB}} \right)k_{iz}h} - {r_{iT}r_{iB}k_{iz}^{2}h^{2}}} \right\}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}} \right\rbrack}} & 15\end{matrix}$

Using the Galerkin technique, as described above, the moment matrix issymmetric. This means that the source subsection and field subsectioncan be swapped and exactly identical couplings result. For example, tocalculate the electromagnetic coupling from current on a volume rooftopto an up-tapered via, simply use equation (10) to calculate the couplingfrom current on an up-tapered via to a volume rooftop. Thus, reciprocityapplies and the resulting method of moments impedance matrix issymmetric.

At step 210, a current distribution in the structure is determinedaccording to the transfer impedance or transfer admittance of eachsubsection and an assumed voltage across each subsection. This can beaccomplished by inverting the Moment Matrix. Now the equation becomes(Big Inverted Moment Matrix)*(Voltage on Each Subsection)=(Current onEach Subsection) and assume all voltages (except where there are inputsto the circuit) are zero. Where the structure is comprised of aconductor, there is zero voltage on all non-input conductors because aperfect conductor must always have zero voltage across it. In otherwords, the Z matrix is inverted such that I=Y*V, where Y=Z⁻¹. The momentmatrix is now an admittance matrix. If the input voltage source isapplied to subsection 1, then V₁ is set to 1 (the value of the appliedvoltage) and the remaining subsection voltages (i.e., all of theremaining v_(x) values) are set to zero. The result is obtained bysubstituting the now known voltages into the voltage vector V and giventhe admittance matrix Y, as calculated above based on the reactionintegral equations, the computer numerically evaluates I=Y*V, themagnitude of the current for every subsection. The actual currentdistribution on each subsection is now the magnitude of the current inthe vector I for each subsection multiplied by the basis function(s)that make up each subsection. The current distribution in the circuit isnow the sum of all the actual current distributions on all thesubsections and the problem is solved. The result is provided to theuser by way of a GUI or other post-processing interface.

Again, the voltages are set to zero when it is assumed that theconductor is a perfect conductor (that is, in the lossless case). If,however, if there is loss in a subsection, then the voltage on thatsubsection will not be zero. Rather, it will be proportional to thecurrent flowing in the subsection. So the final voltage in subsection 1must be R₁*I₁, where the resistance of subsection 1 is R₁, rather thanzero voltage if it were a perfect conductor. So now the equation forsubsection 1 becomes V₁+R₁*I₁=Z₁₁*I₁+Z₁₂*I₂+Z₁₃*I₃+ . . . Since I₁ istypically not known a priori, the R₁*I₁ term is moved over to the righthand side, or: V₁=(Z₁₁−R₁)*I₁+Z₁₂*I₂+Z₁₃*I₃+ . . . In order to make thevoltage on a lossy subsection proportional to the current flowingthrough it as per Ohm's Law, the corresponding Method of Momentsimpedance (Z) matrix is modified by subtracting the Ohmically-generatedvoltage (those voltages resulting from the modeled current flowingthrough a lossy subsection) from the appropriate matrix elements. Aswith all the other reaction integrals, for a Galerkin technique, theelectric field generated by the source subsection is multiplied by thebasis function(s) forming the field subsection before integrating thegenerated electric field over the volume of the subsection. Theprocedure for including loss in the reaction integrals follows exactlythe same procedure as indicated above, however the required integralsare often nearly trivial to perform and are detailed in Appendix 4.

Unlike the electromagnetically induced voltages, the Ohmically inducedvoltages affect only the subsections that the inducing current touches.Thus, if subsection 1 overlaps subsection 2, but is only adjacent tosubsection 3, only Z₁₁ and Z₁₂ are modified (as above), but Z₁₃ is leftalone as I₁ does not induce any Ohmic voltage on subsection 3. In thelossless case, then, the remaining voltages are set to zero; whereas inthe lossy case the remaining voltages are made to be proportional to thecurrent flowing through the subsection as described above.

Conceptually, the above-described computer-implemented method can bedescribed as follows: First, the entire circuit is subsectioned intosmall subsections. Next, a specific pre-determined current distributionis placed on one subsection (i.e., the basis function). That subsectionis then taken all by itself and a voltage is calculated that, say, oneAmp, on that subsection induces at the location of each of the othersubsections (and on itself). This is performed for each subsection inturn. (Mathematically, this fills a matrix with all the pair-wiseelectromagnetic couplings.) Next, a current is placed on all thesubsections at the same time and those currents are adjusted so that thetotal voltage induced on every subsection is zero. (Mathematically, thisis done by matrix inversion.) In other words, a voltage cannot existacross a perfect conductor. To generate current in the circuit in thefirst place, a voltage must be placed on one or more subsections. Thatsubsection can be viewed as the ‘input connector’. The currents on allthe subsections that give us zero voltage on all conductors (except theinput connector) in the circuit form the current distribution on thecircuit and is the solution. If there is loss (i.e., the conductors arenot perfect conductors), the voltage on a conductor is not quite zero.Rather, current flowing in the conductor generates a small voltageacross that conductor. This is just Ohm's Law and is realized by a smallmodification to the equations.

The above reaction integrals all assume the source and field subsectionsare completely within the same layer of dielectric. For a completeanalysis, the reaction integrals between subsections within differentlayers must also be calculated. To do this, source subsection and fieldsubsection are first selected. Then the reaction integral result withboth source and field subsections within the same layer are determinedusing the equations presented above. However, for this example, theactual field subsection is within a different layer. The field layer canbe viewed as being connected to the source layer by a cascade ofrectangular waveguides (the sidewalls of the box and intervening layersof dielectric forming the waveguides). Since the coupling to the fieldsubsection is a sum of waveguide modes, standard waveguide cascadedtransmission line theory can be used to determine the modification ofthe amplitude of each waveguide mode that originates in the source layeronce it arrives in the field layer. The modified waveguide modeamplitudes are then summed to realize the reaction integral for thefield subsection in the field layer due to current in the sourcesubsection in the source layer. This is detailed in Appendix 3.

A short description of the effectiveness of method 200, and,specifically, the use of tapered vias, follows. Presently, as describedabove, when using shielded 3-D multi-layer planar method of moments EM(electromagnetic) analysis, vertical current is modeled using onlyuniform vias. The current everywhere in a given block is verticallydirected and is the same magnitude and phase. In an actual physicalvertical circuit, the vertical current gradually and smoothly changes inboth magnitude and phase as it flows along a long via. In order toaccurately model this using uniform vias, a long via is modeled using astack of short (with respect to wavelength) via blocks. While themagnitude and phase of the current on each block is still constantwithin the block, it can change from one block to the next.

For example, the current distribution on a quarter wave vertical dipoleabove a ground plane, in reality, is a quarter of a sine wave, with zerocurrent at the top (open end) and maximum current at the bottom where itconnects to a groundplane, with the current varying smoothly in between.Since this is a shielded analysis, antenna analysis is problematic.However, a vertical quarter wave open circuited stub is of interest andis considered below. Physically, it is nearly identical to the verticalquarter wave antenna, except for radiation. Just like the antenna, theopen circuited stub has a sinusoidal current distribution.

FIG. 5 shows how this actual sinusoidal current is modeled when usingonly uniform vias. The actual smoothly varying (typically sinusoidal)current for a full half wave dipole or, in a shielded environment, ahalf wave resonator, is shown as curve 502, while the approximationusing the uniform vias is shown as the piecewise-constant function 504,506, 508, 510, 512. The error due to this piecewise-constant functionmodel becomes smaller as shorter via blocks are used to model the opencircuited stub. Accordingly, much finer meshing (i.e., shorter viablocks), and thus, many more subsections and longer analysis time, isrequired to achieve high accuracy.

If tapered vias are added to the uniform vias, with each tapered viaoccupying the exact same via block as each uniform via, bysuperposition, the current in each via block is now modeled as aconstant plus a linear taper. This results in a piecewise-linearrepresentation of the current along the length of the quarter wave longstub, as shown in FIG. 4B, again for a full half wave dipole, or, in ashielded environment, a half wave resonator. (Note that ‘taper’ refersonly to the current density variation within the block. The physicalshape of each via block is still a rectangular prism.) This representsthe actual quarter wave sinusoid much more accurately than uniform viasalone. Thus, for a given accuracy, far fewer subsections are needed.

FIG. 6 shows a vertical open circuited stub. The stub is fed by ahorizontal transmission line in order to verify the successfultransition from horizontal to vertical current, as well as to verify thesubstantial increase in accuracy that tapered vias allow on the verticalportion of the stub. The vertical stub in this figure is divided into2.5 via blocks (the bottom via block is half the length of the other viablocks and is counted as half of a block for speed calculation) asindicated by horizontal lines dividing the length of the vertical stub.FIG. 7 shows results for the various meshing alternatives. Curve 702shows the result when using only long uniform via blocks, 2.5 of them,along the length of the stub. Note that this result is substantiallydifferent from all the other curves. Expecting that the difference isdue to the piecewise-constant approximation (shown in FIG. 5) to theactual sinusoidal current, the circuit is re-meshed with 17 via blocks,each a 10 micron cube. Only uniform vias are used, one per block, so theresult is a much finer piecewise-constant approximation to the actualsinusoidal current. The result of the 17 uniform via blocks is shown ascurve 704, a result much closer to the remaining curves.

To estimate the correct answer, the same open stub is rotated by 90degrees about the horizontal feed line so that the entire open circuitedstub is horizontal and the entire circuit can be analyzed using only thepreviously validated horizontal volume rooftop subsections. This isshown as curve 706. Curve 704 is close but not the same, still bearingsome level of inaccuracy.

The lowest two curves 708 and 710 are nearly identical, one on top ofthe other. One of the curves uses 2.5 via blocks with both uniform andtapered vias. This meshing results in the current on the vertical stubbeing modeled in a piecewise-linear fashion, similar to FIG. 4B. Theother curve shows results using 33 via blocks, each 5 microns long,filled with uniform vias only, which results in a piecewise-constantrepresentation of the current on the stub, similar to FIG. 5.

The analysis results include both the horizontal feedline and thevertical stub. This makes the stub by itself nearly ⅛^(th) of awavelength long at the highest frequency, 175 GHz and 2.5 via blocks per⅛^(th) wavelength gives 20 via blocks per wavelength. Using two viasubsections, one uniform via and one tapered via, per block means thereare 40 subsections per wavelength for a piecewise-linear representationof the current. Using 33 via blocks with uniform vias only to model the⅛^(th) wavelength stub means 264 via blocks per wavelength are employed,with one uniform via subsection for each via block resulting in apiecewise-constant representation of the current. These two meshingsyield nearly the same accuracy. Since both of these results arevirtually identical, and both of these results have higher degrees offreedom (i.e., more subsections) than the horizontal version of thisstub described above, it is reasonable to conclude that these tworesults are the most accurate.

Based on these two final results, we conclude that using uniform plustapered vias allows modelling a vertical circuit with 264/40≈6.6 timesfewer subsections. This means that the matrix solve is now about6.6³≈280 times faster for vertical circuits, as described above. Thisresult will scale independent of circuit or structure size. Inotherwords, one stub, or a thousand stubs will see about the same factorincrease in speed of the main bottleneck in this analysis, the matrixsolve.

In conclusion, the tapered via enables a drastic reduction in analysistime and permits computers to solve previously unsolvable verticalcurrent distribution. The use of tapered vias thus improves thefunctioning of a computer, allowing computers to perform simulationsthat computers were previously unable to perform within reasonable timeconstraints.

Actions associated with implementing all or part of the functions can beperformed by one or more programmable processors executing one or morecomputer programs to perform the computer implemented method. Processorssuitable for the execution of a computer program include, by way ofexample, both general and special purpose, and any one or moreprocessors of any kind of digital computer. Generally, a processor willreceive instructions and data from a read-only memory or a random-accessmemory or both. Components of a computer include a processor forexecuting instructions and one or more memory devices for storinginstructions and data.

While several inventive embodiments have been described and illustratedherein, those of ordinary skill in the art will readily envision avariety of other means and/or structures for performing the functionand/or obtaining the results and/or one or more of the advantagesdescribed herein, and each of such variations and/or modifications isdeemed to be within the scope of the inventive embodiments describedherein. More generally, those skilled in the art will readily appreciatethat all parameters, dimensions, materials, and configurations describedherein are meant to be exemplary and that the actual parameters,dimensions, materials, and/or configurations will depend upon thespecific application or applications for which the inventive teachingsis/are used. Those skilled in the art will recognize, or be able toascertain using no more than routine experimentation, many equivalentsto the specific inventive embodiments described herein. It is,therefore, to be understood that the foregoing embodiments are presentedby way of example only and that, within the scope of the appended claimsand equivalents thereto, inventive embodiments may be practicedotherwise than as specifically described and claimed. Inventiveembodiments of the present disclosure are directed to each individualfeature, system, article, material, and/or method described herein. Inaddition, any combination of two or more such features, systems,articles, materials, and/or methods, if such features, systems,articles, materials, and/or methods are not mutually inconsistent, isincluded within the inventive scope of the present disclosure.

The examples and methods described above are better understood inconnection with the appendices detailed below.

Appendix 1

Basic Variables

Index i is used for summing over all m, n, TE, TM modes.

Index j goes over all dielectric layers, starting with layer j=0 at thetop and increases going down. This index is distinct from j=√{squareroot over (−1)}. Which is in use is determined by context. The zcoordinate starts at zero at the bottom and increases going up.

Wavenumber k=2π/λ=ω√{square root over (με)} in general. Modaladmittances are the ratio of H over E of the standing wave modes, andthus differ by a factor of j from the usual traveling wave modaladmittances.

$\begin{matrix}{k_{iz}^{(j)} = \sqrt{k_{j}^{2} - k_{c}^{2}}} & (1) \\{k_{c}^{2} = {{\left( \frac{m\pi}{a} \right)^{2} + \left( \frac{n\pi}{b} \right)^{2}} = {k_{x}^{2} + k_{y}^{2}}}} & \left( {1a} \right) \\{Y_{i,{TE}}^{(j)} = \frac{k_{iz}^{(j)}}{j\omega\mu_{j}}} & (2) \\{Y_{i,{TM}}^{(j)} = {- \frac{j\omega ɛ_{j}}{k_{iz}^{(j)}}}} & (3)\end{matrix}$

Transverse normalized modal vectors:

$\begin{matrix}{\mspace{79mu}{{e_{iTE} = {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{y}y} \right)}u_{x}\mspace{14mu} m} = 0}},\mspace{20mu}{n > 0},}} & (4) \\{\mspace{79mu}{{= {{{- \sqrt{\frac{2}{ab}}}{\sin\left( {k_{x}x} \right)}u_{y}\mspace{14mu} m} > 0}},\mspace{20mu}{n = 0},\mspace{20mu}{{{otherwise}\mspace{14mu}{for}\mspace{14mu} m\mspace{14mu}{and}\mspace{14mu} n} > {0\text{:}}}}} & (5) \\{= {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{\frac{n}{b}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{x}} - {\frac{m}{a}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (6) \\{\mspace{79mu}{{h_{iTE} = {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{y}y} \right)}u_{y}\mspace{14mu} m} = 0}},\mspace{20mu}{n > 0},}} & (7) \\{\mspace{79mu}{{= {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{x}x} \right)}u_{x}\mspace{14mu} m} > 0}},\mspace{20mu}{n = 0},\mspace{20mu}{{{otherwise}\mspace{14mu}{for}\mspace{14mu} m\mspace{14mu}{and}\mspace{14mu} n} > {0\text{:}}}}} & (8) \\{= {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{\frac{m}{a}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{x}} + {\frac{n}{b}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (9)\end{matrix}$

For the TM modal vectors, both m and n>0:

$\begin{matrix}{e_{iTM} = \begin{matrix}{2{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {\frac{m}{a}{\cos\left( {k_{x}x} \right)}\sin} \right.}} \\{\left. {{\left( {k_{y}y} \right)u_{x}} + {\frac{n}{b}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{y}}} \right\rbrack.}\end{matrix}} & {(10)} \\{= {{N_{2}g_{1}u_{x}} + {N_{1}g_{2}u_{y}}}} & {\left( {10a} \right)} \\{h_{iTM} = \begin{matrix}{2{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{- \frac{n}{b}}{\sin\left( {k_{x}x} \right)}\cos} \right.}} \\{\left. {{\left( {k_{y}y} \right)u_{x}} + {\frac{m}{a}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{y}}} \right\rbrack.}\end{matrix}} & {(11)} \\{= {{{- N_{1}}g_{2}u_{x}} + {N_{2}g_{1}u_{y}}}} & {\left( {11a} \right)}\end{matrix}$ $\begin{matrix}{{{{where}\mspace{14mu} N_{1}} = \frac{2k_{y}}{k_{c}\sqrt{ab}}},{N_{2} = \frac{2k_{x}}{k_{c}\sqrt{ab}}},{g_{1} = {{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}x} \right)}}},{{{and}\mspace{14mu} g_{2}} = {{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}x} \right)}}}} & \left( {11b} \right)\end{matrix}$

Transverse fields for the j^(th) layer and the j^(th) (m, n, TE, TM)mode:

E _(it) ^((j)) ={F _(i) ^((j))sin(k _(iz) ^((j)))+G _(i) ^((j))cos(k_(iz) ^((j)) z)}e _(i),  (12)

H _(it) ^((j)) =−Y _(i) ^((j)) {F _(i) ^((j)))cos(k _(iz) ^((j)) z)−G_(i) ^((j))sin(k _(iz) ^((j)) z)}h _(i).  (13)

F and G are set to match boundary conditions at top and bottom covers.If we change z to c−z, then also change sign of H.

The full TM^(z) field (from pg. 90 of my first notebook, there is noz-component in the TM Magnetic field, the yellow highlighted portion isthe only addition to (12) and (13)):

$\begin{matrix}{i\mspace{14mu}{for}\mspace{14mu}{TM}\mspace{14mu}{modes}\mspace{14mu}{ONL}\text{Y:}} & \; \\{{E_{i}^{(j)} = {{\left\{ {{F_{i}^{(j)}{\sin\left( {k_{iz}^{(j)}z} \right)}} + {G_{i}^{(j)}{\cos\left( {k_{iz}^{(j)}z} \right)}}} \right\} e_{i}} - {k_{iz}^{(j)}N_{3}{g_{3}\left( {x,y} \right)}\left\{ {{F_{i}^{(j)}{\cos\left( {k_{iz}^{(j)}z} \right)}} - {G_{i}^{(j)}{\sin\left( {k_{iz}^{(j)}z} \right)}}} \right\} u_{z}}}},} & (14) \\{H_{i}^{(j)} = {{- Y_{i}^{(j)}}\left\{ {{F_{i}^{(j)}{\cos\left( {k_{iz}^{(j)}z} \right)}} - {G_{i}^{(j)}{\sin\left( {k_{iz}^{(j)}z} \right)}}} \right\}{h_{i}.}}} & (15) \\{{{{where}\mspace{14mu} N_{3}} = \frac{2k_{c}}{\left( k_{iz}^{(j)} \right)^{2}\sqrt{ab}}},{g_{3} = {{\sin\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}}}} & (16)\end{matrix}$

If we change z to (z−c), then also change sign of H and change the signof the z-component of E (which is highlighted in yellow).

When there is only one layer, or all layers have the same dielectric,the j index is not used, which simplifies the notation.

General Representation of Fields in Layered Shielded Media

In general, we use the following form for the z-dependence of alltransverse electric fields in layer k_(iz) when those fields arereferenced to the top cover:

$\begin{matrix}{{{F_{iT}^{(j)}{\sin\left( {k_{iz}^{(j)}\left( {c_{j} - z} \right)} \right)}} + {G_{iT}^{(j)}{\cos\left( {k_{iz}^{(j)}\left( {c_{j} - z} \right)} \right)}}},{or}} & (17) \\{F_{iT}^{(j)}\left\{ {{\sin\left( {k_{iz}^{(j)}\left( {c_{j} - z} \right)} \right)} + {r_{iT}^{(j)}{\cos\left( {k_{iz}^{(j)}\left( {c_{j} - z} \right)} \right)}}} \right\}} & (18) \\{{{wher}\text{e:}\mspace{14mu} r_{iT}^{(j)}} = {\frac{G_{iT}^{(j)}}{F_{iT}^{(j)}}\mspace{14mu}{and}\mspace{14mu} c_{j}\mspace{14mu}{is}\mspace{14mu}{the}\mspace{14mu} z\mspace{14mu}{coordinate}\mspace{14mu}{of}\mspace{14mu}{the}\mspace{14mu}{top}\mspace{14mu}{of}\mspace{14mu}{the}\mspace{14mu} j^{th}\mspace{14mu}{{layer}.}}} & (19)\end{matrix}$

The z-coordinate of the top cover is c₀. The top dielectric layer isLayer 0.

With the above form of the z-dependence used for electric fields, wemust use the following form for magnetic fields:

F _(iT) ^((j))cos(k _(iz) ^((j))(c _(j) −z))−G _(iT) ^((j)))sin(k _(iz)^((j))(c _(j) −z)), or  (20)

F _(iT) ^((j)){cos(k _(iz) ^((j))(c _(j) −z))−r _(iT) ^((j))sin(k _(iz)^((j))(c _(j) −z)}  (21)

For fields referenced to the bottom cover, we substitute a (z−c₁₊₁)dependence, where c_(j+1) is the z-coordinate of the bottom of thej^(th) layer. The notation for the constants changes as follows:

F _(iT) ^((j)) →F _(iB) ^((j)) ,G _(iT) ^((j)) →G _(iB) ^((j)) ,r _(iT)^((j)) →r _(iB) ^((j))  (22)

If there are two layers, N=1, and the layers are numbered 0 and 1. Withlayer numbers going from 0 (at the top) to N (at the bottom, there areN+1 layers and the z-coordinate of the bottom cover, the bottom of thebottom Layer N is c_(N+1)=0=.

To match boundary conditions at the top cover, and to make transversefields continuous across dielectric boundaries for fields referenced tothe top cover, we use the following expressions for F and G (layer 0 isthe top-most layer):

$\begin{matrix}{{F_{iT}^{(0)} = 1},\ {G_{iT}^{(0)} = r_{iT}^{(0)}}} & (23) \\{r_{iT}^{(j)} = {\frac{G_{iT}^{(j)}}{F_{iT}^{(j)}} = {Y_{i}^{(j)}R_{iT}^{(j)}}}} & (24) \\{G_{iT}^{({j + 1})} = {{F_{iT}^{(j)}{\sin\left( {k_{iz}^{(j)}h_{j}} \right)}} + {G_{iT}^{(j)}{\cos\left( {k_{iz}^{(j)}h_{j}} \right)}}}} & (25) \\{F_{iT}^{({j + 1})} = {\frac{Y_{i}^{(j)}}{Y_{i}^{({j + 1})}}\left\lbrack {{F_{iT}^{(j)}{\cos\left( {k_{iz}^{(j)}h_{j}} \right)}} - {G_{iT}^{(j)}{\sin\left( {k_{iz}^{(j)}h_{j}} \right)}}} \right\rbrack}} & (26)\end{matrix}$

The above quantities are used to represent fields starting at the top(0) layer, and working down. For example, for the first two layers ontop, we have:

E _(t) ⁽⁰⁾ =ΣV _(iT){Sin(k _(iz) ⁽⁰⁾(c ₀ −z))+r _(iT) ⁽⁰⁾cos(k _(iz)⁽⁰⁾(c ₀ −z))}e _(i)  (27)

H _(t) ⁽⁰⁾ =ΣV _(iT) Y _(i) ⁽⁰⁾{cos(k _(iz) ⁽⁰⁾(c ₀ −z))−r _(iT)⁽⁰⁾sin(k _(iz) ⁽⁰⁾(c ₀ −z))}h _(i)  (28)

E _(t) ⁽¹⁾ =ΣV _(iT) F _(iT) ⁽¹⁾{sin(k _(iz) ⁽¹⁾(c ₁ −z))+r _(iT)⁽¹⁾cos(k _(iz) ⁽¹⁾(c ₁ −z))}e _(i)  (29)

H _(t) ⁽¹⁾ =ΣV _(iT) Y _(i) ⁽¹⁾{cos(k _(iz) ⁽¹⁾)(c ₁ −z))−r _(iT)⁽¹⁾sin(k _(iz) ⁽¹⁾(c ₁ −z))}h _(i)  (30)

Notice that we meet the impedance boundary condition at the top cover,

${z = c_{0}},{\frac{E_{x}^{(0)}}{H_{y}^{(0)}} = {{- \frac{E_{y}^{(0)}}{H_{x}^{(0)}}} = R_{iT}^{(0)}}},$

and that both electric and magnetic fields are continuous at z=c₁.

If the bottom layer is Layer N, then:

$\begin{matrix}{{F_{iB}^{(N)} = 1},\ {G_{iB}^{(N)} = r_{iB}^{(N)}}} & (31) \\{r_{iB}^{(N)} = {\frac{G_{iT}^{(N)}}{F_{iB}^{(N)}} = {Y_{i}^{(N)}R_{iB}^{(N)}}}} & (32) \\{G_{iB}^{({j - 1})} = {{F_{iB}^{(j)}{\sin\left( {k_{iz}^{(j)}h_{j}} \right)}} + {G_{iB}^{(j)}{\cos\left( {k_{iz}^{(j)}h_{j}} \right)}}}} & (33) \\{F_{iB}^{({j - 1})} = {\frac{Y_{i}^{(j)}}{Y_{i}^{({j - 1})}}\left\lbrack {{F_{iB}^{(j)}{\cos\left( {k_{iz}^{(j)}h_{j}} \right)}} - {G_{iB}^{(j)}{\sin\left( {k_{iz}^{(j)}h_{j}} \right)}}} \right\rbrack}} & (34)\end{matrix}$

The above quantities are used to represent fields starting at thebottom, layer N and working up. (Reminder: If there are two layers, N=1,and the layers are numbered 0 and 1.) For the first two layers on bottom(the physically higher layer listed first), we have:

E _(t) ^((N-1)) =ΣV _(ib) F _(ib) ^((N-1)){sin(k _(iz) ^((N-1))(z−c_(N)))+r _(iB) ^((N-1))cos(k _(iz) ^((N-1))(z−c _(N)))}e _(i)  (35)

H _(t) ^((N-1)) =−ΣV _(ib) Y _(i) ^((N-1)) F _(iB) ^((N-1)){cos(k _(iz)^((N-1))(z−c _(N)))−r _(iB) ^((N-1))sin(k _(iz) ^((N-1))(z−c _(N)))}h_(i)  (36)

E _(t) ^((N)) =ΣV _(iB){sin(k _(iz) ^((N)) z)+r _(iB) ^((N))cos(k _(iz)^((N)) z)}e _(i)  (37)

H _(t) ^((N)) =−ΣV _(iB) Y _(i) ^((N)){cos(k _(iz) ^((N)) z)−r _(iB)^((N))sin(k _(iz) ^((N)) z)}h _(i)  (38)

Notice that we meet the impedance boundary condition at the bottom cover

${z = 0},{{- \frac{E_{x}^{(N)}}{H_{y}^{(N)}}} = {\frac{E_{y}^{(N)}}{H_{x}^{(N)}} = R_{iB}^{(N)}}}$

and that both electric and magnetic fields are continuous at z=c_(N).

Layer 0 is at the top, and Layer N is at the bottom of the dielectricstack. The top of Layer j is at z=c_(j) and is referred to as Levelj.The bottom of Layer j is at z=c_(j+1) referred to as Level (j+1). Thethickness of Layer j is h_(j)=c_(j)−_(j+1). Thus, the top cover is Level0, and the next level down is Level 1. The bottom cover is Level N+1 andis located at z=0. If there are two dielectric layers, then N=1 and wehave Layers 0 and 1, and Levels 0, 1, and 2. Note: In Sonnet, forhistorical reasons, level numbers are one less than what is used here.The Layer numbers are the same.

Thus, a level takes the number of the layer directly below it. Often, werepresent the fields above a given level, j, with equations of the formof (29) and (30), and the fields below with equations of the form of(35) and (36). In this case, for the fields above Level j we have (39)and (40) with modal coefficients V_(iT), and for the fields below Levelj we have (41) and (42) with modal coefficients V_(iB):

E _(t) ^((j-1)) =ΣV _(it) F _(iT) ^((j-1)){sin(k _(iz) ^((j-1))(c _(j-1)−z))+r _(iT) ^((j-1))cos(k _(iz) ^((j-1))(c _(j-1) −z))}e _(i)  (39)

H _(t) ^((j-1)) =ΣV _(iT) Y _(i) ^((j-1)) F _(iT) ^((j-1)){cos(k _(iz)^((j-1))(c _(j-1) −z))−r _(iT) ^((j-1))sin(k _(ix) ^((j-1))(c _(j-1)−z))}h _(i)   (40)

E _(t) ^((j)) =ΣV _(iB) F _(iB) ^((j)){sin(K _(iz) ^((j))(z−c _(j+i)))+r_(iB) ^((j))cos(k _(iz) ^((j))(z−c _(j+1)))}e _(i)  (41)

H _(t) ⁽¹⁾ =ΣV _(iB) Y _(i) ^((j)) F _(iB) ^((j)){cos(k _(iz) ^((j))(z−c_(j+1)))−r _(iB) ^((j))sin(k _(iz) ^((j))(z—c _(j+1)))}h _(i)  (42)

For a complete description of the fields in the entire structure, weneed to make sure that the electric-fields are continuous across levelj, thus we set (39) equal to (41) at z=c_(j), and then use (25) and(33), yielding:

V _(iT) F _(iT) ^((j−1){sin(k) _(iz) ^((j−1)) h _(j-1))+r _(iT)^((j-1))cos(k _(iz) ^((j-1)) h _(j-1))}=V _(iB) F _(iB) ^((j)){sin(k_(iz) ^((j)) h _(j))+r _(iB) ^((j))cos(k _(iz) ^((j)) h _(j))}V _(iT) G_(iT) ^((j)) =V _(iB) G _(iB) ^((j-1))  (44)

We can use the same weighting coefficients, V_(i), for both summationsif we select:

$\begin{matrix}{V_{i} = {{V_{iT}G_{iT}^{(j)}} = {V_{iB}G_{iB}^{({j - 1})}}}} & (45) \\{V_{iT} = {{\frac{V_{i}}{G_{iT}^{(j)}}\mspace{14mu}{and}\mspace{14mu} V_{iB}} = \frac{V_{i}}{G_{iB}^{({j - 1})}}}} & (46)\end{matrix}$

This gives us the fields above, Layer (j−1), and below Level j, Layer(j) as:

$\begin{matrix}{E_{t}^{({j - 1})} = {\sum{\frac{V_{i}}{G_{iT}^{(j)}}F_{iT}^{({j - 1})}\left\{ {{\sin\left( {k_{zi}^{({j - 1})}\left( {c_{j - 1} - z} \right)} \right)} + {r_{iT}^{({j - 1})}{\cos\left( {k_{iz}^{({j - 1})}\left( {c_{j - 1} - z} \right)} \right)}}} \right\} e_{i}}}} & (47) \\{H_{t}^{({j - 1})} = {\sum{\frac{V_{i}}{G_{iT}^{(j)}}Y_{i}^{({j - 1})}F_{iT}^{({j - 1})}\left\{ {{\cos\left( {k_{iz}^{({j - 1})}\left( {c_{j - 1} - z} \right)} \right)} - {r_{iT}^{({j - 1})}{\sin\left( {k_{iz}^{({j - 1})}\left( {c_{j - 1} - z} \right)} \right)}}} \right\} h_{i}}}} & (48) \\{E_{t}^{(j)} = {\sum{\frac{V_{i}}{G_{iB}^{({j - 1})}}F_{iB}^{(j)}\left\{ {{\sin\left( {k_{iz}^{(j)}\left( {z - c_{j + 1}} \right)} \right)} + {r_{iB}^{(j)}{\cos\left( {k_{iz}^{(j)}\left( {z - c_{j + 1}} \right)} \right)}}} \right\} e_{i}}}} & (49) \\{H_{t}^{(j)} = {- {\sum{\frac{V_{i}}{G_{iB}^{({j - 1})}}Y_{i}^{(j)}F_{iB}^{(j)}\left\{ {{\cos\left( {k_{iz}^{(j)}\left( {z - c_{j + 1}} \right)} \right)} - {r_{iB}^{(j)}{\sin\left( {k_{iz}^{(j)}\left( {z - c_{j + 1}} \right)} \right)}}} \right\} h_{i}}}}} & (50)\end{matrix}$

Now of interest is the discontinuity in tangential magnetic field atLevel j, z=c_(j), between Layers j−1 and j:

$\begin{matrix}{\left\lbrack {H_{t}^{({j - 1})} - H_{t}^{(j)}} \right\rbrack_{z = c_{j}} = {{\sum{\frac{V_{i}}{G_{iT}^{(j)}}Y_{i}^{({j - 1})}F_{iT}^{({j - 1})}\left\{ {{\cos\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)} - {r_{iT}^{({j - 1})}{\sin\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)}}} \right\} h_{i}}} + {+ {\sum{\frac{V_{i}}{G_{iB}^{({j - 1})}}Y_{i}^{(j)}F_{iB}^{(j)}\left\{ {{\cos\left( {k_{iz}^{(j)}h_{j}} \right)} - {r_{iB}^{(j)}{\sin\left( {k_{iz}^{(j)}h_{j}} \right)}}} \right\} h_{i}}}}}} & (51)\end{matrix}$

Using (26) and (34), we can now define Ŷ_(i) ^((j)):

$\begin{matrix}{\left\lbrack {H_{t}^{({j - 1})} - H_{t}^{(j)}} \right\rbrack_{z = c_{j}} = {\sum{{V_{i}\left( {{Y_{i}^{(j)}\frac{F_{iT}^{(j)}}{G_{iT}^{(j)}}} + {Y_{i}^{({j - 1})}\frac{F_{iB}^{({j - 1})}}{G_{iB}^{({j - 1})}}}} \right)}h_{i}}}} & {(52)} \\{= {\sum{V_{i}{\overset{\hat{}}{Y}}_{i}^{(j)}h_{i}}}} & {(53)}\end{matrix}$ $\begin{matrix}{{{w{here}}\mspace{14mu}{\overset{\hat{}}{Y}}_{i}^{(j)}} = {\left( {{Y_{i}^{(j)}\frac{F_{iT}^{(j)}}{G_{iT}^{(j)}}} + {Y_{i}^{({j - 1})}\frac{F_{iB}^{({j - 1})}}{G_{iB}^{({j - 1})}}}} \right) = \left( {\frac{1}{R_{iT}^{(j)}} + \frac{1}{R_{iB}^{({j - 1})}}} \right)}} & (54)\end{matrix}$

Physically, Ŷ_(i) ^((j)) is the admittance of the top cover rotatedthrough the layers above Level j to Level j and connected in parallelwith the admittance of the bottom cover rotated through the layers belowLevel j to Level j. Note that while the j in the waveguidecharacteristic admittance, Ŷ_(i) ^((j)), refers to Layer j, the j inŶ_(i) ^((j)) refers to Level j.

For j=0 or N+1 it might appear that values are needed for Y_(i) ⁽⁻¹⁾ orY_(i) ^((N+1)), which are in non-existent layers. However, in all suchcases, the non-existent characteristic admittances cancel with the samequantity in F_(iB) ⁽⁻¹⁾, (34), or F_(iT) ^((N+1)) (26). Numerically,those characteristic admittances can be set to any arbitrary value aslong as they are not set to zero.

Typically, the discontinuity in tangential magnetic field at Level j(caused by the source current) is set equal to (53) and this determinesthe V_(i). Then (39)-(42) determine the fields in all layers.

Fields Surrounding an Original Via in a Three-Layered Medium

For the new vias, we start with three layers and place a via only on themiddle layer, illustrated in FIG. 8. Then we take the limit of theresulting fields as the thickness of the middle layer goes to zero. Thisis the Green's function for z-directed current in a layered medium. Wethen use that Green's function to characterize a uniform and a taperedvia.

From this point forward, we assume that the dielectric of all threesubstrates is the same. In other words, k_(iz) ^((j))=k_(iz) and Y_(i)^((j))=Y_(i) for all j.

For the R variables, a T subscript means the impedance of the top cover(R_(T) ⁽⁰⁾) transformed down the layer stack by the interveningwaveguides to the indicated level. The number in the parentheses in thesuperscript refers to the dielectric layer. The B subscript means thatit is the impedance of the bottom cover (R_(B) ⁽²⁾) transformed up thelayer stack to the indicated level.

Because the exciting source is a z-directed current, all fields are TMonly, see (29), (30), (37), (38) in the ViasOriginal document. Asdescribed in that document, the via fields are represented by twoweighted summations. The first summation, weighted by V_(ia), representsfields with a discontinuity in the tangential magnetic field at Level 1.This marks the top end of the via. The second summation, weighted byV_(ib), provides the same discontinuity with opposite sign at Level 2.This marks the bottom end of the via. Adapting (47)-(50) with j=1 forthe V_(ia)summation j=2 for the V_(ib) summation, the transverse fieldsfor each layer in the above figure are:

$\begin{matrix}{E_{t}^{(0)} = {\sum\limits_{TM}{\left( {\frac{V_{ia}}{G_{iT}^{(1)}} - \frac{V_{ib}}{G_{iT}^{(2)}}} \right)\left\{ {{\sin\left( {k_{iz}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}\left( {c_{0} - z} \right)} \right)}}} \right\} e_{i}}}} & (55) \\{H_{t}^{(0)} = {\sum\limits_{TM}{\left( {\frac{V_{ia}}{G_{iT}^{(1)}} - \frac{V_{ib}}{G_{iT}^{(2)}}} \right)Y_{i}\left\{ {{\cos\left( {k_{iz}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}\left( {c_{0} - z} \right)} \right)}}} \right\} h_{i}}}} & (56) \\{E_{t}^{(1)} = {\sum\limits_{TM}{\left( {{\frac{V_{i\alpha}}{G_{iB}^{(0)}}F_{iB}^{(1)}\left\{ {{\sin\left( {k_{iz}\left( {z - c_{2}} \right)} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}\left( {z - c_{2}} \right)} \right)}}} \right\}} - {\frac{V_{ib}}{G_{iT}^{(2)}}F_{iT}^{(1)}\left\{ {{\sin\left( {k_{iz}\left( {c_{1} - z} \right)} \right)} + {r_{iT}^{(1)}{\cos\left( {k_{iz}\left( {c_{1} - z} \right)} \right)}}} \right\}}} \right)e_{i}}}} & (57) \\{H_{t}^{(1)} = {\sum\limits_{TM}{\begin{pmatrix}\begin{matrix}{{{F_{R}\left( {\Delta x} \right)}{F_{R}({\Delta y})}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}} - {\frac{V_{ia}}{G_{iB}^{(0)}}Y_{i}F_{iB}^{(1)}}} \\\left\{ {{\cos\left( {k_{iz}\left( {z - c_{2}} \right)} \right)} - {r_{iB}^{(1)}{\sin\left( {k_{iz}\left( {z - c_{2}} \right)} \right)}}} \right\}\end{matrix} \\{{- \frac{V_{ib}}{G_{iT}^{(2)}}}Y_{i}F_{iT}^{(1)}\left\{ {{\cos\left( {k_{iz}\left( {c_{1} - z} \right)} \right)} - {r_{iT}^{(1)}{\sin\left( {k_{iz}\left( {c_{1} - z} \right)} \right)}}} \right\}}\end{pmatrix}h_{i}}}} & (58) \\{E_{t}^{(2)} = {\sum\limits_{TM}{\left( {\frac{V_{ia}}{G_{iB}^{(0)}} - \frac{V_{ib}}{G_{iB}^{(1)}}} \right)\left\{ {{\sin\left( {k_{iz}z} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}z} \right)}}} \right\} e_{i}}}} & (59) \\{H_{t}^{(2)} = {- {\sum\limits_{TM}{\left( {\frac{V_{ia}}{G_{iB}^{(0)}} - \frac{V_{ib}}{G_{iB}^{(1)}}} \right)Y_{i}\left\{ {{\cos\left( {k_{iz}z} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}z} \right)}}} \right\} h_{i}}}}} & (60)\end{matrix}$

In (58), the minus sign in front of the V_(ia) summation is because ofthe (z−c₂) dependence. The minus sign in front of the V_(ib) summationis because we are subtracting that summation from the V_(ia) summation.All fields are proportional to the current on the via, assumed above tobe unity.

For the above fields, there is a discontinuity in transverse magneticfield in the V_(ia) summation at z=c₁ and a discontinuity in thetransverse magnetic field in the V_(ib) summation at z=c₂. The totaltransverse magnetic field also includes the transverse magnetic fielddue to the via current in Layer 1, i.e., the first term of (58), see(27) in the ViasOriginal document. Since the total magnetic field mustbe continuous across both Level 1 and Level 2, we solve for the V_(ia)that make the field continuous across Level 1 at z=c₁ and for the V_(ib)to make the field continuous across Level 2 at z=c₂. See, for example(30) in the ViasOriginal document. Using (53):

$\begin{matrix}{\left\lbrack {H_{t}^{(0)} - H_{t}^{(1)}} \right\rbrack_{z = c_{1}} = {{\sum\limits_{TM}{\left( {{V_{ia}{\overset{\hat{}}{Y}}_{i}^{(1)}} - {{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}} \right)h_{i}}} = 0}} & (61) \\{V_{ia} = {{\overset{\hat{}}{Z}}_{i}^{(1)}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}} & (62) \\{\left\lbrack {H_{t}^{(1)} - H_{t}^{(2)}} \right\rbrack_{z = c_{2}} = {{- {\sum\limits_{TM}{\left( {{V_{ib}{\overset{\hat{}}{Y}}_{i}^{(2)}} - {{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}} \right)h_{i}}}} = 0}} & (63) \\{V_{ib} = {{\overset{\hat{}}{Z}}_{i}^{(2)}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}} & (64)\end{matrix}$

For determining the system matrix coefficients, we are interested onlyin the electric field. Substituting (62) and (64) into (55), (57), and(59), we obtain the transverse electric field:

$\begin{matrix}{E_{t}^{(0)} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)\left\{ {{\sin\left( {k_{iz}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}\left( {c_{0} - z} \right)} \right)}}} \right\} e_{i}}}} & (65) \\{E_{t}^{(1)} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {{\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}}F_{iB}^{(1)}\left\{ {{\sin\left( {k_{iz}\left( {z - c_{2}} \right)} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}\left( {z - c_{2}} \right)} \right)}}} \right\}} - {\frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}F_{iT}^{(1)}\left\{ {{\sin\left( {k_{iz}\left( {c_{1} - z} \right)} \right)} + {r_{iT}^{(1)}{\cos\left( {k_{iz}\left( {c_{1} - z} \right)} \right)}}} \right\}}} \right)e_{i}}}} & (66) \\{E_{t}^{(2)} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)\left\{ {{\sin\left( {k_{iz}z} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}z} \right)}}} \right\} e_{i}}}} & (67)\end{matrix}$

Checking for continuous transverse electric field across both dielectricinterfaces, using (25) and (33):

$\begin{matrix}{\left\lbrack E_{t}^{(0)} \right\rbrack_{z = c_{1}} = {\left\lbrack E_{t}^{(1)} \right\rbrack_{z = c_{1}} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {{\overset{\hat{}}{Z}}_{i}^{(1)} - {{\overset{\hat{}}{Z}}_{i}^{(2)}\frac{G_{iT}^{(1)}}{G_{iT}^{(2)}}}} \right)e_{i}}}}} & (68) \\{\left\lbrack E_{t}^{(1)} \right\rbrack_{z = c_{2}} = {\left\lbrack E_{t}^{(2)} \right\rbrack_{z = c_{2}} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {{{\overset{\hat{}}{Z}}_{i}^{(1)}\frac{G_{iB}^{(1)}}{G_{iB}^{(0)}}} - {\overset{\hat{}}{Z}}_{i}^{(2)}} \right)e_{i}}}}} & (69)\end{matrix}$

Note that (68) is consistent with (54) in the ViasOriginal document, andthat (69) is consistent with (51) in the ViasOriginal document.

We also need the vertical, z-directed, electric fields due to the via.Using (14):

$\begin{matrix}{E_{z}^{(0)} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)\left\{ {{\cos\left( {k_{iz}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}\left( {c_{0} - z} \right)} \right)}}} \right\} u_{z}}}} & (70) \\{E_{z}^{(1)} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta\; y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {{k_{iz}N_{3}^{2}\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}}F_{iB}^{(1)}\left\{ {{\cos\left( {k_{iz}\left( {z - c_{2}} \right)} \right)} - {r_{iB}^{(1)}{\sin\left( {k_{iz}\left( {z - c_{2}} \right)} \right)}}} \right\}} + {N_{3}^{2}k_{iz}\frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}F_{iT}^{(1)}\left\{ {{\cos\left( {k_{iz}\left( {c_{1} - z} \right)} \right)} - {r_{iT}^{(1)}\sin\left( {k_{iz}\left( {c_{1} - z} \right)} \right)}} \right\}} - {\frac{4}{ab}\frac{j\;\omega\;\mu}{k_{iz}^{2}}}} \right)u_{z}}}}} & (71) \\{E_{z}^{(2)} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)\left\{ {{\cos\left( {k_{iz}z} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}z} \right)}}} \right\} u_{z}}}}} & (72)\end{matrix}$

As mentioned in the text following (14), E normally (has a leadingnegative sign for a (z−c) dependence and a leading positive sign for a(c−z) dependence. In addition, the {circumflex over (Z)}_(i) ⁽²⁾summation (which causes the via current to stop at Level 2) issubtracted from the {circumflex over (Z)}_(i) ⁽¹⁾ summation (whichcauses the via current to start at Level 1). The combination of thesetwo requirements determines the signs of (70)-(72).

The final term in (71) is from (26) of the ViasOriginal document. Thisis the z-directed electric field generated by the via itself. Itcorresponds to the transverse magnetic field of (27) in the ViasOriginaldocument, which is what generates the discontinuity in the totaltransverse magnetic field used in (61)-(64), above, to determine themodal amplitudes, V_(ia) and V_(ib).

Following is a list of all the F and G constants used above, written outaccording to (23)-(26) and (31)-(34). These equations are needed furtheron in this document.

G _(iT) ⁽¹⁾=sin(k _(iz) h ₀)+r _(iT) ⁽⁰⁾cos(k _(iz) h ₀)  (73)

F _(i) ^(T)(1)=cos(k _(iz) h ₀)−r _(iT) ⁽⁰⁾sin(k _(iz) h ₀)  (74)

G _(iT) ⁽²⁾ =F _(iT) ⁽¹⁾sin(k _(iz) h ₁)+G _(iT) ⁽¹⁾cos(k _(iz) h_(j1))  (75)

F _(iT) ⁽²⁾=cos(k _(iz) h ₁)−G _(iT) ⁽¹⁾sin(k _(iz) h ₁)  (76)

G _(iB) ⁽¹⁾=sin(k _(iz) h ₂)+r _(iB) ⁽²⁾cos(k _(iz) h ₂)  (77)

F _(iB) ⁽¹⁾=cos(k _(iz) h ₂)−r _(iB) ²)sin(k _(iz) h ₂)  (78)

G _(iB) ⁽⁰⁾ =F _(iB) ⁽¹⁾sin(k _(iz) h ₁)+G _(iB) ⁽¹⁾cos(k _(iz) h₁)  (79)

F _(iB) ⁽⁰⁾ =F _(iB) ⁽¹⁾cos(k _(iz) h ₁)−G _(iB) ⁽¹⁾sin(k _(iz) h₁)  (80)

In order to determine if the via creates a discontinuity in thez-directed electric field, we need to evaluate the electric field,(70)-(72), on each side of Level 1 and Level 2. (73)-(80) are used tosimplify the expressions.

$\begin{matrix}{\left\lbrack E_{z}^{(0)} \right\rbrack_{z = c_{1}} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)F_{iT}^{(1)}u_{z}}}} & (81) \\{\left\lbrack E_{z}^{(1)} \right\rbrack_{z = c_{1}} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {{k_{iz}N_{3}^{2}\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}}F_{iB}^{(0)}} + {k_{iz}N_{3}^{2}\frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}F_{iT}^{(1)}} - {\frac{4}{ab}\frac{j\;{\omega\mu}}{k_{iz}^{2}}}} \right)u_{z}}}}} & (82) \\{\left\lbrack E_{z}^{(1)} \right\rbrack_{z = c_{2}} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {{k_{iz}N_{3}^{2}\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}}F_{iB}^{(1)}} + {k_{iz}N_{3}^{2}\frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}F_{iT}^{(2)}} - {\frac{4}{ab}\frac{j\;{\omega\mu}}{k_{iz}^{2}}}} \right)u_{z}}}}} & (83) \\{\left\lbrack E_{z}^{(2)} \right\rbrack_{z = c_{2}} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)F_{iB}^{(1)}u_{z}}}}} & (84)\end{matrix}$

Evaluating the discontinuity in z-directed electric field across Level 1and then applying (54):

$\begin{matrix}{\left\lbrack {E_{z}^{(0)} - E_{z}^{(1)}} \right\rbrack_{z = c_{1}} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}}} & {(85)} \\{\left\{ {{k_{iz}N_{3}^{2}{{\overset{\hat{}}{Z}}_{i}^{(1)}\left( {\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}} + \frac{F_{iB}^{(0)}}{G_{iB}^{(0)}}} \right)}} - {\frac{4}{ab}\frac{j\;\omega\;\mu}{k_{iz}^{2}}}} \right\} u_{z}} & \\{= {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}}} & {{~~~~~~~~}(86)} \\{\left\{ {\frac{k_{iz}N_{3}^{2}}{Y_{i}} - {\frac{4}{ab}\frac{j\omega\mu}{k_{iz}^{2}}}} \right\} u_{z}} & \\{= {\sum\limits_{TM}\frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}\end{matrix}}{Y_{i}}}} & {\left( {86a} \right)} \\{\left\{ {1 - \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right\} u_{z}} & \end{matrix}$

Evaluating the discontinuity in z-directed electric field across Level 2and then applying (54):

$\begin{matrix}{\left\lbrack {E_{z}^{(1)} - E_{z}^{(2)}} \right\rbrack_{z = c_{2}} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}}} & {(87)} \\{\left\{ {{{- k_{iz}}N_{3}^{2}{{\overset{\hat{}}{Z}}_{i}^{(2)}\left( {\frac{F_{iT}^{(2)}}{G_{iT}^{(2)}} + \frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}} \right)}} + {\frac{4}{ab}\frac{j\omega\mu}{k_{iz}^{2}}}} \right\} u_{z}} & \\{= {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}}} & {(88)} \\{\left\{ {{- \frac{k_{iz}N_{3}^{2}}{Y_{i}}} + {\frac{4}{ab}\frac{j\omega\mu}{k_{iz}^{2}}}} \right\} u_{\; z}} & \\{= {\sum\limits_{TM}\frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}\end{matrix}}{Y_{i}}}} & {\left( {88a} \right)} \\{\left\{ {\frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}} - 1} \right\} u_{z}} & \end{matrix}$

(86a) and (88a) are an alternative way to write (86) and (88) and areobtained by using (16) and noting that:

$\begin{matrix}{{\left( {\frac{4}{ab}\frac{j\omega\mu}{k_{iz}^{2}}} \right)\left( \frac{Y_{i}}{k_{iz}N_{3}^{2}} \right)} = {{\left( {\frac{4}{ab}\frac{j\omega\mu}{k_{iz}^{2}}} \right)\left( \frac{Y_{i}}{k_{iz}} \right)\left( \frac{k_{iz}^{4}ab}{4k_{c}^{2}} \right)} = \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}}} & \left( {88b} \right)\end{matrix}$

Note that the discontinuity in the electric field at both levels isindependent of the via layer thickness. Also, if we take the limit as h₁goes to zero of (81) and (84), we find that both

[E_(z)⁽⁰⁾]_(z = c₁)  and  [E_(z)⁽²⁾]_(z = c₂)

go to zero linearly with h₁, see (100) and (112), below. This means thatz-directed electric field in the via layer, Layer 1, is equal to (88)when the via layer is infinitesimally thin. When we complete the Green'sfunction, with an infinitesimally thin via layer, the effect of thisz-directed electric field is included as an impulse of electric field atthe level of the source current, z=z′, (122).

To evaluate the electric fields in the layer above the via, (65) and(70), per via length, as the length of the via, h₁, goes to zero, weneed to evaluate:

$\begin{matrix}{\lim\limits_{h_{1}\rightarrow 0}{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)}} & (89)\end{matrix}$

Evaluating the inverse of the first term of (89) in the parenthesesusing (54), (79), and (80):

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(1)}G_{iT}^{(1)}} = {{Y_{i}\left( {\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}} + \frac{F_{iB}^{(0)}}{G_{iB}^{(0)}}} \right)}G_{iT}^{(1)}}} & {(90)} \\{= {Y_{i}\left( {F_{iT}^{(1)} + \frac{F_{iB}^{(0)}G_{iT}^{(1)}}{G_{iB}^{(0)}}} \right)}} & {(91)} \\{= {Y_{i}\left( {F_{iT}^{(1)} + {\frac{\left\lbrack {{F_{iB}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}} - {G_{iB}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}}} \right\rbrack}{{F_{iB}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}} + {G_{iB}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}}}G_{iT}^{(1)}}} \right)}} & {{~~~~~~~~~~}(92)}\end{matrix}$

Substituting one term of the Taylor series for each sine and cosine in(92), we have:

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(1)}G_{iT}^{(1)}} \cong {Y_{i}\left( {F_{iT}^{(1)} + {\frac{F_{iB}^{(1)} - {G_{iB}^{(1)}k_{iz}h_{1}}}{{F_{iB}^{(1)}k_{iz}h_{1}} + G_{iB}^{(1)}}G_{iT}^{(1)}}} \right)}} & (93)\end{matrix}$

Evaluating the inverse of the second term of (89) using (54), (75), and(76):

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(2)}G_{iT}^{(2)}} = {{Y_{i}\left( {\frac{F_{iT}^{(2)}}{G_{iT}^{(2)}} + \frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}} \right)}G_{iT}^{(2)}}} & {(94)} \\{= {Y_{i}\left( {F_{iT}^{(2)} + \frac{F_{iB}^{(1)}G_{iT}^{(2)}}{G_{iB}^{(1)}}} \right)}} & {(95)} \\{= {Y_{i}\left( {{F_{iT}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}} - {G_{iT}^{(1)}\sin\left( {k_{iz}h_{1}} \right)} +} \right.}} & {{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}(96)} \\\left. {\frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}\left\lbrack {{F_{iT}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}} + {G_{iT}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}}} \right\rbrack} \right) & \end{matrix}$

Substituting one term of the Taylor series for each sine and cosine in(96), we have:

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(2)}G_{iT}^{(2)}} \cong {Y_{i}\left( {F_{iT}^{(1)} - {G_{iT}^{(1)}k_{iz}h_{1}} + {\frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}\left\lbrack {{F_{iT}^{(1)}k_{iz}h_{1}} + G_{iT}^{(1)}} \right\rbrack}} \right)}} & (97)\end{matrix}$

Inverting (93) and (97) and then subtracting:

$\begin{matrix}{{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)} \cong {\frac{1}{h_{1}Y_{i}}\left( {\frac{1}{F_{iT}^{(1)} + {\frac{\begin{matrix}{F_{iB}^{(1)} -} \\{G_{iB}^{(1)}k_{iz}h_{1}}\end{matrix}}{\begin{matrix}{{F_{iB}^{(1)}k_{iz}h_{1}} +} \\G_{iB}^{(1)}\end{matrix}}G_{iT}^{(1)}}} -} \right.}} & {(98)} \\\left. \frac{1}{\begin{matrix}{F_{iT}^{(1)} - {G_{iT}^{(1)}k_{iz}h_{1}} +} \\{\frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}\left\lbrack {{F_{iT}^{(1)}k_{iz}h_{1}} + G_{iT}^{(1)}} \right\rbrack}\end{matrix}} \right) & \\{\cong {\frac{1}{h_{1}Y_{i}}\left( {\frac{{F_{iB}^{(1)}k_{iz}h_{1}} + G_{iB}^{(1)}}{\begin{matrix}\begin{matrix}\left( {{F_{iB}^{(1)}k_{iz}h_{1}} + G_{iB}^{(1)}} \right) \\{F_{iT}^{(1)} +}\end{matrix} \\{\begin{pmatrix}{F_{iB}^{(1)} -} \\{G_{iB}^{(1)}k_{iz}h_{1}}\end{pmatrix}G_{iT}^{(1)}}\end{matrix}} -} \right.}} & {{~~~~~~~~~~~~~~~~~~~~~~~}(99)} \\\left. \frac{G_{iB}^{(1)}}{\begin{matrix}{{G_{iB}^{(1)}F_{iT}^{(1)}} - {G_{iB}^{(1)}G_{iT}^{(1)}k_{iz}h_{1}} + F_{iB}^{(1)}} \\\left( {{F_{iT}^{(1)}k_{iz}h_{1}} + G_{iT}^{(1)}} \right)\end{matrix}} \right) & \end{matrix}$

Noting that the denominators of both fractions are equal, we can easilysubtract them, and then take the appropriate limit, yielding:

$\begin{matrix}{{\lim\limits_{h_{1}\rightarrow 0}{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iT}^{(1)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iT}^{(2)}}} \right)}} = {\frac{1}{Y_{i}}\left( \frac{F_{iB}^{(1)}k_{iz}}{{G_{iB}^{(1)}F_{iT}^{(1)}} + {G_{iT}^{(1)}F_{iB}^{(1)}}} \right)}} & (100)\end{matrix}$

Now, in a manner very similar to (89)-(100) for the fields above thevia, we evaluate the electric fields in the layer below the via, (67)and (72), per via length, as the length of the via, h₁, goes to zero, weneed to evaluate:

$\begin{matrix}{\lim\limits_{h_{1}\rightarrow 0}{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)}} & (101)\end{matrix}$

Evaluating the inverse of the first term in the parentheses of (101)using (54), (79), and (80) and assuming the V₁ are the same for each ofthe dielectric layers (i.e., all three dielectric layers have the samedielectric):

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(1)}G_{iB}^{(0)}} = {{Y_{i}\left( {\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}} + \frac{F_{iB}^{(0)}}{G_{iB}^{(0)}}} \right)}G_{iB}^{(0)}}} & {{~~~~~~~~~~~~~~~~~~~}(102)} \\{= {Y_{i}\left( {\frac{F_{iT}^{(1)}G_{iB}^{(0)}}{G_{iT}^{(1)}} + F_{iB}^{(0)}} \right)}} & {(103)} \\{= {Y_{i}\left( {{\left( {{F_{iB}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}} + {G_{iB}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}}} \right)\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}}} +} \right.}} & {(104)} \\\left. {{F_{iB}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}} - {G_{iB}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}}} \right) & \end{matrix}$

Substituting one term of the Taylor series for each sine and cosine in(104), we have:

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(1)}G_{iB}^{(0)}} \cong {Y_{i}\left( {{\left( {{F_{iB}^{(1)}k_{iz}h_{1}} + G_{iB}^{(1)}} \right)\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}}} + F_{iB}^{(1)} - {G_{iB}^{(1)}k_{iz}h_{1}}} \right)}} & (105)\end{matrix}$

Evaluating the inverse of the second term in parentheses of (101) using(54), (75), and (76) and assuming all the Y₁ are equal (i.e., all threedielectric layers have the same dielectric):

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(2)}G_{iB}^{(1)}} = {{Y_{i}\left( {\frac{F_{iT}^{(2)}}{G_{iT}^{(2)}} + \frac{F_{iB}^{(1)}}{G_{iB}^{(1)}}} \right)}G_{iB}^{(1)}}} & {(106)} \\{= {Y_{i}\left( {\frac{F_{iT}^{(2)}G_{iB}^{(1)}}{G_{iT}^{(2)}} + F_{iB}^{(1)}} \right)}} & {(107)} \\{{= {Y_{i}\left( {\frac{\begin{bmatrix}{{F_{iT}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}} -} \\{G_{iT}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}}\end{bmatrix}G_{iB}^{(1)}}{\begin{matrix}{{F_{iT}^{(1)}{\sin\left( {k_{iz}h_{1}} \right)}} +} \\{G_{iT}^{(1)}{\cos\left( {k_{iz}h_{1}} \right)}}\end{matrix}} + F_{iB}^{(1)}} \right)}}\mspace{169mu}} & {(108)}\end{matrix}$

Substituting one term of the Taylor series for each sine and cosine in(108), we have:

$\begin{matrix}{{{\overset{\hat{}}{Y}}_{i}^{(2)}G_{iB}^{(1)}} \cong {Y_{i}\left( {\frac{\left\lbrack {F_{iT}^{(1)} - {G_{iT}^{(1)}k_{z,i}^{(1)}h_{1}}} \right\rbrack G_{iB}^{(1)}}{{F_{iT}^{(1)}k_{iz}h_{1}} + G_{iT}^{(1)}} + F_{iB}^{(1)}} \right)}} & (109)\end{matrix}$

Inverting (105) and (109) and then subtracting:

$\begin{matrix}{{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)} \cong {\frac{1}{h_{1}Y_{i}}\left( {\frac{1}{\begin{matrix}\begin{matrix}\begin{pmatrix}{{F_{iB}^{(1)}k_{iz}h_{1}} +} \\G_{iB}^{(1)}\end{pmatrix} \\{\frac{F_{iT}^{(1)}}{G_{iT}^{(1)}} +}\end{matrix} \\\begin{matrix}{F_{iB}^{(1)} -} \\{G_{iB}^{(1)}k_{iz}h_{1}}\end{matrix}\end{matrix}} - \frac{1}{\frac{\begin{bmatrix}{F_{iT}^{(1)} -} \\{G_{iT}^{(1)}k_{iz}h_{1}}\end{bmatrix}G_{iB}^{(1)}}{\begin{matrix}{{F_{iT}^{(1)}k_{iz}h_{1}} +} \\G_{iT}^{(1)}\end{matrix}} + F_{iB}^{(1)}}} \right)}} & {(110)} \\{\cong {\frac{1}{h_{1}Y_{i}}\left( {\frac{G_{iT}^{(1)}}{\begin{matrix}\begin{pmatrix}{{F_{iB}^{(1)}k_{tz}h_{1}} +} \\G_{iB}^{(1)}\end{pmatrix} \\\begin{matrix}{F_{iT}^{(1)} + {G_{iT}^{(1)}F_{iB}^{(1)}} -} \\{G_{iB}^{(1)}G_{iT}^{(1)}k_{iz}h_{1}}\end{matrix}\end{matrix}} - \frac{{F_{iT}^{(1)}k_{iz}^{(1)}h_{1}} + G_{iT}^{(1)}}{\begin{matrix}\begin{matrix}\begin{bmatrix}{F_{iT}^{(1)} -} \\{G_{iT}^{(1)}k_{iz}h_{1}}\end{bmatrix} \\\begin{matrix}{G_{iB}^{(1)} +} \\F_{iB}^{(1)}\end{matrix}\end{matrix} \\\begin{pmatrix}{{F_{iT}^{(1)}k_{iz}h_{1}} +} \\G_{iT}^{(1)}\end{pmatrix}\end{matrix}}} \right)}} & {(111)}\end{matrix}$

Noting that the denominators of both fractions are equal, we can easilysubtract them, and then take the appropriate limit, yielding:

$\begin{matrix}{{\lim\limits_{h_{1}\rightarrow 0}{\frac{1}{h_{1}}\left( {\frac{{\overset{\hat{}}{Z}}_{i}^{(1)}}{G_{iB}^{(0)}} - \frac{{\overset{\hat{}}{Z}}_{i}^{(2)}}{G_{iB}^{(1)}}} \right)}} = {{- \frac{1}{Y_{i}}}\left( \frac{F_{iT}^{(1)}k_{z,i}^{(1)}}{{G_{iB}^{(1)}F_{iT}^{(1)}} + {G_{iT}^{(1)}F_{iB}^{(1)}}} \right)}} & (112)\end{matrix}$

The denominator of both (100) and (112) are the same and must beexpanded before we arrive at the desired Green's function. Using (77),(74), (73), and (78) in that order:

$\begin{matrix}{{{G_{iB}^{(1)}F_{iT}^{(1)}} + {G_{iT}^{(1)}F_{iB}^{(1)}}}==\left( {{\sin\left( {k_{iz}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}h_{2}} \right)}}} \right)} & {{~~~~~~~~~~~~}(113)} \\{\left( {{\cos\left( {k_{iz}h_{0}} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}h_{0}} \right)}}} \right) +} & \\{\left( {{\sin\left( {k_{iz}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}h_{0}} \right)}}} \right)} & \\{\left( {{\cos\left( {k_{iz}h_{2}} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}h_{2}} \right)}}} \right)} & \\{= {{{\sin\left( {k_{iz}h_{2}} \right)}{\cos\left( {k_{iz}h_{0}} \right)}} -}} & {(114)} \\{{r_{iT}^{(0)}{\sin\left( {k_{iz}h_{2}} \right)}{\sin\left( {k_{iz}h_{0}} \right)}} +} & \\{{r_{iB}^{(2)}{\cos\left( {k_{iz}h_{2}} \right)}{\cos\left( {k_{iz}h_{0}} \right)}} -} & \\{{r_{iB}^{(2)}r_{iT}^{(0)}{\cos\left( {k_{iz}h_{2}} \right)}{\sin\left( {k_{iz}h_{0}} \right)}} +} & \\{{{\sin\left( {k_{iz}h_{0}} \right)}{\cos\left( {k_{iz}h_{2}} \right)}} -} & \\{{r_{iB}^{(2)}{\sin\left( {k_{iz}h_{0}} \right)}{\sin\left( {k_{iz}h_{2}} \right)}} +} & \\{{r_{iT}^{(0)}{\cos\left( {k_{iz}h_{0}} \right)}{\cos\left( {k_{iz}h_{2}} \right)}} -} & \\{r_{iB}^{(2)}r_{iT}^{(0)}{\cos\left( {k_{iz}h_{0}} \right)}{\sin\left( {k_{iz}h_{2}} \right)}} & \end{matrix}\mspace{31mu}$

This simplifies considerably if we apply the equations for the sine andcosine of the sum of two angles (ID-1) and (ID-2) in theUsefulIdentities document and let h=h₀+h₂.

$\begin{matrix}{{{G_{iB}^{(1)}F_{iT}^{(1)}} + {G_{iT}^{(1)}F_{iB}^{(1)}}} = {{\sin\left( {k_{iz}h} \right)} + {r_{iT}^{(0)}\cos\left( {k_{iz}h} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}h} \right)}} -}} & {(115)} \\{r_{iB}^{(2)}r_{iT}^{(0)}{\sin\left( {k_{iz}h} \right)}} & \\{= {{\left( {1 - {r_{iB}^{(2)}r_{iT}^{(0)}}} \right){\sin\left( {k_{iz}h} \right)}} +}} & {(116)} \\{\left( {r_{iT}^{(0)} + r_{iB}^{(2)}} \right){\cos\left( {k_{iz}h} \right)}} & \end{matrix}$

We now define two impedances, one for Top (Z_(T)) and a second forBottom (Z_(B)) electric fields. Note that the same impedance is used forboth transverse fields and for z-directed fields. We will also definethe z-coordinate of the, now infinitesimally thin Layer 1, the vialayer, to be z′, which makes h₀=h−z′ and h₂=z′. Substituting (116) into(100) and into (112) and expanding the numerator using (78) and (74) wehave:

$\begin{matrix}{{Z_{T}\left( z^{\prime} \right)} = {\frac{k_{iz}}{Y_{i}}\frac{{\cos\left( {k_{iz}z^{\prime}} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}z^{\prime}} \right)}}}{{\left( {1 - {r_{iB}^{(2)}r_{iT}^{(0)}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT}^{(0)} + r_{iB}^{(2)}} \right){\cos\left( {k_{iz}h} \right)}}}}} & (117) \\{{Z_{B}\left( z^{\prime} \right)} = {{- \frac{k_{iz}}{Y_{i}}}\frac{{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}}{{\left( {1 - {r_{iB}^{(2)}r_{iT}^{(0)}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT}^{(0)} + r_{iB}^{(2)}} \right){\cos\left( {k_{iz}h} \right)}}}}} & (118)\end{matrix}$

Substituting (117) into (65) and (118) into (67), we have the transverseelectric field for an infinitesimally short z-directed rectangular2Δx×2Δy cylinder of current located at z=z′ and centered at (x₀, y₀):

$\begin{matrix}{{{E_{tT} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\mspace{40mu}{Z_{T}\left( z^{\prime} \right)}\left\{ {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right\} e_{i}}}},\mspace{79mu}{z \geq {z'}}}\mspace{59mu}} & (119) \\{{E_{tB} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}}}\mspace{59mu}{{{Z_{B}\left( z^{\prime} \right)}\left\{ {{\sin\left( {k_{iz}z} \right)} + {r_{iB}^{(2)}\cos\left( {k_{iz}z} \right)}} \right\} e_{i}},\mspace{79mu}{z \leq {z'}}}} & (120)\end{matrix}$

Both the transverse electric and magnetic fields are continuous acrossthe original Level 1 and Level 2 boundaries, and thus, there is nodiscontinuity or singularity at z=z′.

Substituting (117) into (70) and (118) into (72), we have the z-directedelectric field for an infinitesimally short z-directed rectangular2Δx×2Δy cylinder of current located at z=z′ and centered at (xn, vn):

$\begin{matrix}{{E_{zT} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}}}\mspace{95mu}{{{Z_{T}\left( z^{\prime} \right)}\left\{ {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right\} u_{z}},\mspace{79mu}{z > {z'}}}} & (121) \\{\mspace{79mu}{{E_{zV}-={\sum\limits_{TM}{\frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}\end{matrix}}{Y_{i}}\left\{ {1 - \frac{j\omega\mu k_{iz}^{2}Y_{i}}{k_{c}^{2}}} \right\} u_{z}}}},{z = z^{\prime}}}} & (122) \\{{E_{zB} = {- {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}}}}}\mspace{40mu}{{{g_{3}\left( {x,y} \right)}{Z_{B}\left( z^{\prime} \right)}\left\{ {{\cos\left( {k_{iz}z} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}z} \right)}}} \right\} u_{z}},\mspace{79mu}{z < {z'}}}} & (123)\end{matrix}$

(122) is taken from (88a), it could also have been taken from (88), andrepresents an impulse, or singularity, of the electric field at thelevel of the infinitesimally short via.

For the remainder of this work, we eliminate the layer-indicatingsuperscripts for the normalized cover resistances. We are concernedhenceforth only with r_(iT) ⁽⁰⁾ and r_(iB) ⁽²⁾ which are below referredto simply as r_(IT) and r_(iB).

For the work below, we need to concentrate on the z-dependence of theabove equations, which may be re-written with all the constants andextraneous dependencies combined into a single variable (the subscriptG=Green's function, T=Top, B=Bottom, t=transverse, z=z):

$\begin{matrix}\begin{matrix}{\mspace{25mu}{T_{it} = {\frac{k_{iz}}{Y_{i}}\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}}}} & {(124)} \\{T_{iz} = {\frac{k_{iz}^{2}}{Y_{i}}\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\{\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}\end{matrix}}}} & {(125)} \\{= {T_{it}k_{iz}N_{3}{g_{3}\left( {x,y} \right)}}} & \\{= \frac{T_{iv}k_{iz}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} & {\left( {125a} \right)} \\{T_{iv} = {k_{iz}\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{Y_{i}}}} & {\left( {125b} \right)} \\{E_{GtT} = {\sum\limits_{TM}{{T_{it}\left( {{\cos\left( {k_{iz}z^{\prime}} \right)} - {r_{iB}\sin\left( {k_{iz}z^{\prime}} \right)}} \right)}\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} +} \right.}}} & {(126)} \\{{\left. {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)e_{i}},{z \geq z^{\prime}}} & \\{E_{GtB} = {- {\sum\limits_{TM}{T_{it}\left( {{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)} -}\mspace{45mu} \right.}}}} & {(127)} \\{\left. {r_{iT}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} \right)\left( {{\sin\left( {k_{iz}z} \right)} +} \right.} & \\{{\left. {r_{iB}{\cos\left( {k_{iz}z} \right)}} \right)e_{i}},\;{z \leq z^{\prime}}} & \\{E_{GzT} = {\sum\limits_{TM}{{T_{iz}\left( {{\cos\left( {k_{iz}z^{\prime}} \right)} - {r_{iB}\sin\left( {k_{iz}z^{\prime}} \right)}} \right)}\left( {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} -} \right.}}} & {(128)} \\{{\left. {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)u_{z}},{z > z^{\prime}}} & \\{{E_{GzV} = {- {\sum\limits_{TM}{T_{iv}\left\{ {1 - \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right\} u_{z}}}}},{z = z^{\prime}}} & {(129)} \\{E_{GzB} = {\sum\limits_{TM}{T_{iz}\left( {{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)} - {r_{iT}\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} \right)}}} & {{~~~~~~~}(130)} \\{{\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)u_{z}},{z < z^{\prime}}} & \end{matrix} & \;\end{matrix}$

Note that the change of sign for (127) and (130) is due to the minussign in (118). All fields (125)-(130), except (129), actually go to zeroas the via length goes to zero. That is why we present the fields interms of per via length as the via length goes to zero. Thus, theyrepresent the derivative of the electric field with respect to the vialength as the via length goes to zero. See, for example, (100) and(112). It is interesting that (128) and (130) are equal when evaluatedat z=z′ while (126) and (127) are not. This suggests that the presenceof an x-y sheet of z-directed current forces a discontinuity in thederivative with respect to z of the transverse electric field.

Transverse Fields of the New Uniform Via

For the transverse fields of the new uniform via at a given level, z,due to current flowing in the via below the level z, we integrate (126)from z′=0 to z. This gives us the total field at z that is due tocurrent on the via that is below z. Integrating the i^(th) term andletting E_(i)e_(i), =E₁:

$\begin{matrix}{{\int_{z^{\prime} = 0}^{z}{E_{iGtT}{dz}^{\prime}}} = {T_{it}\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)}} & {(131)} \\{{\int_{z^{\prime} = 0}^{z}{\left( {{\cos\left( {k_{iz}z^{\prime}} \right)} - {r_{iB}{\sin\left( {k_{iz}z^{\prime}} \right)}}} \right){dz}^{\prime}}} =} & \\{= {\frac{T_{it}}{k_{iz}}\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)}} & {(132)} \\{\left\lbrack {{\sin\left( {k_{iz}z^{\prime}} \right)} + {r_{iB}{\cos\left( {k_{iz}z^{\prime}} \right)}}} \right\rbrack_{z^{\prime} = 0}^{z} =} & \\{= {\frac{T_{it}}{k_{iz}}\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)}} & {{~~~~~~~~~~~}(133)} \\{\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}} - r_{iB}^{(2)}} \right)} & \end{matrix}$

Multiplying out the last line, above, and moving the

$\frac{T_{it}}{k_{iz}}$

term to the LHS tor now tor convenience, we must be careful to rememberto move it back for the final result):

$\begin{matrix}{{\frac{k_{iz}}{T_{it}}{\int_{z^{l} = 0}^{z}{E_{iGtT}d\; z^{\prime}}}} = {{{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {r_{iB}r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (134)\end{matrix}$

For the transverse fields of the new uniform via at a given level, z,due to current flowing in the via above the level z, we integrate (127)from z′=z to h. This gives us the total field at z that is due tocurrent on the via that is above z.

$\begin{matrix}{{\int_{z^{\prime} = z}^{h}{E_{iGtB}d\; z^{\prime}}} = {T_{it}\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} & {(135)} \\{\int_{z^{\prime} = z}^{h}\left( {{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)} -} \right.} & \\{{\left. {r_{iT}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} \right)d\; z^{\prime}} =} & \\{= {\frac{T_{it}}{k_{iz}}\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} & {(136)} \\{\left\lbrack {{- {\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} - {r_{iT}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}} \right\rbrack_{z^{\prime} = z}^{h} =} & \\{= {\frac{T_{it}}{k_{iz}}\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}\cos\left( {k_{iz}z} \right)}} \right)}} & {(137)} \\{\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} - r_{iT} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)} & \end{matrix}$

Multiplying out the last line, above and moving the

$\frac{T_{it}}{k_{iz}}$

term to the LHS tor now (tor convenience, we must be careful to rememberto move it back for the final result):

$\begin{matrix}{{\frac{k_{iz}}{T_{it}}{\int_{z^{\prime} = z}^{h}{E_{iGtB}{dz}^{\prime}}}} = {{{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}{\sin\left( {k_{iz}z} \right)}{{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}++}r_{iB}{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iB}r_{iT}{\cos\left( {k_{iz}z} \right)}} + {r_{iB}r_{iT}^{(0)}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (138)\end{matrix}$

The total transverse field at a given level, z, is the sum of the fielddue to via current from below level z added to the via current fromabove level z. In other words, we subtract (138) (because of the minussign leading (127)) from (134) and noting that most terms cancel, whatwe have left is:

$\begin{matrix}{{\frac{k_{iz}}{T_{it}}E_{itUV}} = {{r_{iT}{\sin\left( {k_{iz}z} \right)}} - {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}r_{iT}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (139) \\{{\frac{k_{iz}}{T_{tr}}E_{itUV}} = {{r_{iT}{\sin\left( {k_{iz}z} \right)}} - {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}{r_{iT}\left( {{\cos\left( {k_{iz}z} \right)} - {\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)}}}} & (140) \\{E_{tUV} = {\sum\limits_{TM}{\frac{T_{it}}{k_{iz}}\left\{ {{r_{iT}{\sin\left( {k_{iz}z} \right)}} - {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}{r_{iT}\left( {{\cos\left( {k_{iz}z} \right)} - {\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)}}} \right\} e_{i}}}} & (141)\end{matrix}$

Reorganizing terms, we note a similarity with the transverse fields fora tapered via, (189a):

$\begin{matrix}{{E_{tUV} = {\sum\limits_{TM}{\frac{T_{it}}{k_{iz}}\left\{ {{- {r_{iB}\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)}} + {r_{iT}\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} \right\} e_{i}}}}\ } & \left( {141a} \right)\end{matrix}$

Coupling Between the New Uniform Via and Surface Rooftops

For coupling between the new Uniform Via centered at (x₀, y₀) andx-directed surface rooftops centered at (x₁, y_(i), z), we must use(141) to evaluate:

S _(UVtoRFX) =∫E _(tUV)(z)·J _(RFX)dxdy  (142)

For coupling to rooftops located at the bottom of the uniform via, weevaluate E_(tUV) at z=0:

${\begin{matrix}{\left\lbrack E_{tUV} \right\rbrack_{z = 0} = {\underset{TM}{\Sigma}\frac{T_{it}}{k_{iz}}\left( {{{- r_{iB}}{\sin\left( {k_{iz}h} \right)}} - {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)e_{i}}} & \left( {142a} \right) \\{S_{UVtoRFXbot} = {\underset{TM}{\Sigma}{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}\frac{T_{it}}{k_{iz}}{\left( {{{- r_{iB}}{\sin\left( {k_{iz}h} \right)}} - {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)\left\lbrack {{e_{TM}\left( {x_{0},y_{0}} \right)} \cdot u_{x}} \right\rbrack}}} & (143)\end{matrix}S_{UVtoRFXbot}} = {\underset{TM}{\Sigma}\frac{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{2}{g_{1}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( \frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}} \\\left( {{{- r_{iB}}{\sin\left( {k_{iz}h} \right)}} - {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}$

This result is exactly identical to the result for the old uniform via.In the ViasOriginal document, see Appendix 2.

Likewise, coupling between a New Uniform Via and a y-directed surfacerooftop subsection located at the lower end of the via is (changes from(144) noted in yellow):

$S_{UVtoRFYbot} = {\underset{TM}{\Sigma}\frac{{F_{R}\left( {\Delta x} \right)}{F_{T}\left( {\Delta y} \right)}N_{1}{g_{2}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( \frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}} \\\left( {{{- r_{iB}}{\sin\left( {k_{iz}h} \right)}} - {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}$

For coupling between the new uniform via centered at (x₀, y₀) andsurface rooftops centered at (x₁, y_(i)) on the upper end of the via, weevaluate (141) at z=h.

$\begin{matrix}{\left\lbrack E_{tUV} \right\rbrack_{z = h} = {\sum\limits_{TM}{\frac{T_{iz}}{k_{iz}}\left( {{r_{iT}^{(2)}{\sin\left( {k_{iz}h} \right)}} + {r_{iB}^{(2)}{r_{iT}^{(0)}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)e_{i}}}} & (147) \\{\mspace{40mu}{{S_{UVtoRFXtop} = {\sum\limits_{TM}{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}\frac{T_{it}}{k_{iz}}{\left( {{r_{iT}^{(0)}{\sin\left( {k_{iz}h} \right)}} + {r_{iB}^{(2)}r_{iT}^{(0)}\mspace{191mu}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}} \right)\left\lbrack {{e_{TM}\left( {x_{0},y_{0}} \right)} \cdot u_{x}} \right\rbrack}}}}{S_{UVtoRFXtop} = {\sum\limits_{TM}{\frac{N_{2}{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}{g_{1}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( \frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {{r_{iT}{\sin\left( {k_{iz}h} \right)}} +} \right.} \\\left. {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}} \right)\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}}}}} & (148)\end{matrix}$

Likewise, coupling between a New Uniform Via and a y-directed surfacerooftop subsection located at the upper end of the via is:

$S_{UVtoRFYtop} = {\sum\limits_{TM}{\frac{N_{1}{F_{R}\left( {\Delta x} \right)}{F_{T}\left( {\Delta y} \right)}{g_{2}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( \frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}\left( {{r_{iT}{\sin\left( {k_{iz}h} \right)}} + {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}}} \right)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}}$

Coupling Between the New Uniform Via and Volume Rooftops

For coupling to the volume vias, we must integrate (141) over the lengthof the via. Extracting the z-dependent portion for integration:

$\begin{matrix}{\int_{0}^{h}{\left( {{r_{iT}{\sin\left( {k_{iz}z} \right)}} - {r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}{r_{iB}\left( {{\cos\left( {k_{iz}z} \right)} - {\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)}}} \right){dz}}} & (151) \\{= {{\frac{1}{k_{iz}}\left\lbrack {{{- r_{iT}}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}{r_{iB}\left( {{\sin\left( {k_{iz}z} \right)} + {\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} \right)}}} \right\rbrack}\begin{matrix}\;^{h} & \; \\\;_{z = 0} & \;\end{matrix}}} & (152) \\{= {\frac{1}{k_{iz}}\left( {{{- r_{iT}}{\cos\left( {k_{iz}h} \right)}} + r_{iT} - r_{iB} + {r_{iB}^{(2)}{\cos\left( {k_{iz}h} \right)}} + {r_{iT}r_{iB}\left( {{\sin\left( {k_{iz}h} \right)} - {\sin\left( {k_{iz}h} \right)}} \right)}} \right)}} & (153) \\{= {\frac{1}{k_{iz}}\left( {r_{iT} - r_{iB}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)}} & (154)\end{matrix}$

Substituting (154) into the curly braced portion of (141) and expandingT_(it), we have the tangential electric field integrated vertically overthe length of the substrate:

$\begin{matrix}{{\int_{0}^{h}{E_{tUV}dz}} = {\sum\limits_{TM}{\frac{T_{it}}{k_{iz}^{2}}\left( {r_{iT} - r_{iB}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)e_{i}}}} & (155) \\{S_{VRFXtoUV} = {\sum\limits_{TM}{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}\frac{T_{it}}{k_{iz}^{2}}\left( {r_{iT} - r_{iB}} \right){\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)\left\lbrack {{e_{TM}\left( {x_{0},y_{0}} \right)} \cdot u_{x}} \right\rbrack}}}} & (156) \\{S_{UVtoVRFX} = {\Sigma_{TM}\frac{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{2}{g_{1}\left( {x_{1},y_{1}} \right)}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}}\left( \frac{\left( {r_{iT} - r_{iB}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}} & (157)\end{matrix}$

Note that this result for the new uniform via is identical to the resultfor the original uniform via as is shown in (A2-4) through (A2-6) inAppendix 2 of the OriginalVia documents.

Note that (88) in the VolumeRooftops document, with (90), (27) and (10a)of the same document substituted in, is identical to (157), above. Thismeans that the coupling from a new Uniform Via (UV) to a Volume RFX(VRFX) is the same as the coupling from a VRFX to a UV. This indicatesthat reciprocity holds, a good reality check on the correctness of bothderivations.

The UV to VRFY follows with minor changes (highlighted) to (157):

$S_{UVtoVRFY} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{T}\left( {\Delta y} \right)}N_{1}{g_{2}\left( {x_{1},y_{1}} \right)}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}}\left( \frac{\left( {r_{iT} - r_{iB}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)}}$

This result is equivalent to (88) in the VolumeRooftops document. Inthat document, note that the k_(iz) ² in the first denominator becomesk_(iz) when it cancels with the k_(ix) in C_(i)(x₁, y₁), (90), thusmatching this result.

Normal (z-Directed) Fields of the New Uniform Via

For the normal fields of the new uniform via at a given level, z, due tocurrent flowing in the via below the level z, we integrate (128) fromz′=0 to z. This gives us the total field at z that is due to current onthe via that is below z. Integrating the i^(th) term and lettingE_(i)u_(z), =E_(i):

$\begin{matrix}{{\int_{z^{\prime} = 0}^{z}{E_{iGzT}{dz}^{\prime}}} = {{T_{iz}\left( {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)}{\int_{z^{\prime} = 0}^{z}{\left( {{\cos\left( {k_{iz}z^{\prime}} \right)} - {r_{iB}{\sin\left( {k_{iz}z^{\prime}} \right)}}} \right){dz}^{\prime}}}}} & (159) \\{\mspace{140mu}{= {\frac{T_{iz}}{k_{iz}}{{\left( {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)\left\lbrack {{\sin\left( {k_{iz}z^{\prime}} \right)} + {r_{iB}{\cos\left( {k_{iz}z^{\prime}} \right)}}} \right\rbrack}^{z}\mspace{11mu}}_{z^{\prime} = 0}}}\;} & (160) \\{\mspace{155mu}{= {\frac{T_{iz}}{k_{iz}}\left( {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}} - r_{iB}^{(2)}} \right)}}} & (161)\end{matrix}$

Multiplying out the last line, above, and moving the

$\frac{T_{iz}}{k_{iz}}$

term to the LHS for now (for convenience, we must be careful to rememberto move it back for the final result):

$\begin{matrix}{{\frac{k_{iz}}{T_{iz}}{\int_{z^{\prime} = 0}^{z}{E_{iGzT}{dz}^{\prime}}}} = {{{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {r_{iB}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {r_{iT}r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} + {r_{iT}r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (162)\end{matrix}$

For the normal fields of the new uniform via at a given level, z, due tocurrent flowing in the via above the level z, we integrate (130) fromz′=z to h. This gives us the total field at z that is due to current onthe via that is above z.

$\begin{matrix}{{\int_{z^{\prime} = z}^{h}{E_{iGzB}{dz}^{\prime}}} = {{{T_{iz}\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)}{\int_{z^{\prime} = z}^{h}{\left( {{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}} \right){dz}^{\prime}}}} =}} & (163) \\{= {\frac{T_{iz}}{k_{iz}}{\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)\left\lbrack {{- {\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} - {r_{iT}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}} \right\rbrack}_{z^{\prime} = z}^{h}}} & (164) \\{= {\frac{T_{iz}}{k_{iz}}\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)\left( {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} - r_{iT} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)}} & (165)\end{matrix}$

Multiplying out the last line, above and moving the

$\frac{T_{iz}}{k_{iz}}$

term to the LHS for now (for convenience, we must be careful to rememberto move it back for the final result):

$\begin{matrix}{{\frac{k_{iz}}{T_{iz}}{\int_{z^{\prime} = z}^{h}{E_{iGzB}{dz}^{\prime}}}} = {{{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\cos\left( {k_{iz}z} \right)}} + {r_{iT}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iB}{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}{\sin\left( {k_{iz}z} \right)}} - {r_{iT}r_{iB}{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (166)\end{matrix}$

The total normal field at a given level, z, is (129) added to the fielddue to via current from below level z added to the via current fromabove level z. We start by adding (162) and (166) and make use of thesum of two angles identities, ID-1 and ID-2:

$\begin{matrix}{{{\frac{k_{iz}}{T_{iz}}\left( {{\int_{z^{\prime} = 0}^{z}{E_{iGzT}{dz}^{\prime}}} + {\int_{z^{\prime} = z}^{h}{E_{iGzB}{dz}^{\prime}}}} \right)} = {{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iB}^{(2)} + r_{iT}^{(0)}} \right){\cos\left( {k_{iz}h} \right)}} + {Z_{1}(z)}}}\mspace{79mu}{where}} & (167) \\{{Z_{1}(z)} = {{r_{iT}r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}{\sin\left( {k_{iz}z} \right)}} - {r_{iB}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\cos\left( {k_{iz}z} \right)}}}} & (168)\end{matrix}$

Or, reorganizing terms of (168) in a manner similar to (141a), we alsohave:

Z ₁(z)=r _(iB)(r _(iT) sin(k _(iz)(h−z))−cos(k _(iz)(h−z)))+r _(iT)(r_(iB) sin(k _(iz) z)−cos(k _(iz) z))  (168a)

Reforming the summation over all TM modes of (128)-(130), and including(129), we have the z-directed field for the new uniform via:

$\begin{matrix}{E_{zUV} = {\sum\limits_{TM}{\left\{ {{T_{iz}\frac{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\\begin{matrix}{{\left( {r_{iB} + r_{iT}} \right){\cos\left( {k_{iz}h} \right)}} +} \\{Z_{1}(z)}\end{matrix}\end{matrix}}{k_{iz}}} - {T_{iv}\left( {1 - \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right)}} \right\} u_{z}}}} & (169)\end{matrix}$

Substituting in (125a), which puts T_(iz), in terms of T_(iv), andnoting the similarity between the denominator of (125) and the numeratorof the first fraction in (169), we have:

$\begin{matrix}{{E_{zUV} = {\sum\limits_{TM}{T_{iv}\left\{ {1 + \frac{Z_{1}(z)}{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\{\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}\end{matrix}} - 1 + \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right\} u_{z}}}}\ } & \left( {169a} \right)\end{matrix}$

Noting that two terms cancel, we substitute in (125b) for T_(iv):

$\begin{matrix}{E_{zUV} = {\sum\limits_{TM}{k_{iz}\frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}\end{matrix}}{Y_{i}}\left\{ {\frac{Z_{1}(z)}{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\{\left( {r_{iB} + r_{iT}} \right){\cos\left( {k_{iz}h} \right)}}\end{matrix}} + \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right\} u_{z}}}} & \left( {169b} \right)\end{matrix}$

New Uniform Via Self-coupling

For this, we must integrate (169b) over the thickness of the substrate.Starting with the Z₁(z) portion:

$\begin{matrix}{{\int_{0}^{h}{{Z_{1}(z)}{dz}}} = {{\int_{0}^{h}{r_{iT}r_{iB}\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}{\sin\left( {k_{iz}z} \right)}} -}} & {(170)} \\{{r_{iB}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\cos\left( {k_{iz}z} \right)}{dz}}} & \\{= {\frac{1}{k_{iz}}\left\lbrack {{r_{iT}r_{iB}\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}r_{iB}{\cos\left( {k_{iz}z} \right)}} +} \right.}} & {(171)} \\\left. {{r_{iB}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{\sin\left( {k_{iz}z} \right)}}} \right\rbrack_{z = 0}^{h} & \\{= {\frac{1}{k_{iz}}\left( {{r_{iT}r_{iB}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} - {r_{iT}r_{iB}}} \right.}} & {(172)} \\\left. {\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right) - {r_{iB}{\sin\left( {k_{iz}h} \right)}} - {r_{iT}{\sin\left( {k_{iz}h} \right)}}} \right) & \\{= {\frac{1}{k_{iz}}\left( {{2r_{iT}r_{iB}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} - {\left( {r_{iT} + r_{iB}} \right){\sin\left( {k_{iz}h} \right)}}} \right)}} & {(173)}\end{matrix}$

Integrating (169b):

$\begin{matrix}{{\int_{0}^{h}{{E_{zUV} \cdot u_{z}}{dz}}} = {\sum\limits_{TM}{k_{iz}\frac{\begin{matrix}{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}k_{iz}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}\end{matrix}}{Y_{i}}\left( {\frac{\int_{0}^{h}{{Z_{1}(z)}dz}}{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\{\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}\end{matrix}} + {\frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}h}} \right)}}} & \left( {173a} \right)\end{matrix}$

Performing the Z₁(z) integration by using (173), and bringing the k_(iz)term inside the parentheses, and evaluating at the center of the fieldvia subsection at (x₁, y_(i)) and multiplying by the Fouriercoefficients of the field via subsection, we have the new Uniform Via tonew Uniform Via system matrix element:

$\begin{matrix}{S_{UVtoUV} = {\sum\limits_{TM}{\frac{\begin{matrix}{{F_{R}^{2}\left( {\Delta x} \right)}{F_{R}^{2}\left( {\Delta y} \right)}N_{3}^{2}} \\{{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}\end{matrix}}{Y_{i}}\left( {\frac{\begin{matrix}{{2r_{iT}{r_{iB}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)}} -} \\{\left( {r_{iT} + r_{iB}} \right){\sin\left( {k_{iz}h} \right)}}\end{matrix}}{\begin{matrix}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} +} \\{\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}\end{matrix}} + {\frac{j\omega\mu k_{iz}^{2}Y_{i}}{k_{c}^{2}}h}} \right)}}} & (174)\end{matrix}$

Note that this is identical to (60) with (A3-4) substituted in, in theViasOriginal document. Reference (88b), in this document, to see how thefinal term matches with the final term in (60) in the ViasOriginaldocuments. Thus, the original uniform via couples to other originaluniform vias in exactly the same way as the new uniform via couples tonew uniform vias. Since this is the same conclusion as was reached forcoupling to both surface and volume subsections, we conclude that theoriginal uniform via and the new uniform via are identical.

Transverse Fields for the Tapered Via

The current on the Tapered Via (TV) basis function has maximum currentat the top end of the via and tapers down to zero at the bottom end ofthe via:

$\begin{matrix}{J_{zTV} = \frac{z}{h}} & (175)\end{matrix}$

As with the new Uniform Via transverse field derivation, above, we mustintegrate (126) from z′=0 to z in order to get the field at a given zdue to current below z. Only now, (126) is multiplied by (175). E_(iGtT)is the i^(th) term of the summation representing the Green's function.For convenience, we let E_(iGtT)=E_(wGtT)e_(i), and we work withE_(iGtT) in what follows.

$\begin{matrix}{{\frac{1}{T_{it}}{\int_{0}^{z}{E_{iGtT}{dz}^{\prime}}}} = {{\frac{\begin{matrix}{{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} +} \\{r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}\end{matrix}}{h}{\int_{0}^{Z}{z^{\prime}{\cos\left( {k_{iz}z^{\prime}} \right)}}}} - {r_{iB}z^{\prime}{\sin\left( {k_{iz}z^{\prime}} \right)}{dz}^{\prime}}}} & (176)\end{matrix}$

Using identities ID-8a and ID-9a in the UsefullIdentities documents:

$\begin{matrix}{\;{= {\frac{\begin{matrix}{{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} +} \\{r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}\end{matrix}}{h}\left\lbrack {{\frac{z^{\prime}}{k_{iz}}{\sin\left( {k_{iz}z^{\prime}} \right)}} + {\frac{1}{k_{iz}^{2}}{\cos\left( {k_{iz}z^{\prime}} \right)}} +} \right.}}} & {(177)} \\\left. {{\frac{r_{iB}z^{\prime}}{k_{iz}}{\cos\left( {k_{iz}z^{\prime}} \right)}} - {\frac{r_{iB}}{k_{iz}^{2}}{\sin\left( {k_{iz}z^{\prime}} \right)}}} \right\rbrack_{0}^{z} & \\{= {\frac{\begin{matrix}{{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} +} \\{r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}\end{matrix}}{k_{iz}h}\left( {{z{\sin\left( {k_{iz}z} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} - \frac{1}{k_{iz}} +} \right.}} & {{~~~~~~~~~~~~~~~~~}(178)} \\\left. {{r_{iB}z{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}}} \right) & \end{matrix}$

Bringing the leading RHS denominator over to the LHS and thenmultiplying (178) out, term by term:

$\begin{matrix}{{\frac{k_{iz}h}{T_{it}}{\int_{0}^{z}{E_{iGtT}d\; z^{\prime}}}} = {{z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {\frac{1}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{1}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iT}r_{iB}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}}}} & (179)\end{matrix}$

We must also calculate the field at z due to via current above z′. Forthis, we integrate (127) from z′=z to h in order to get the field at zdue to current above z. As above, (127) is multiplied by (175). Theminus sign on the LHS of (180) is where we are storing the minus signleading the RHS of (127).

$\begin{matrix}{{{- \frac{1}{T_{it}}}{\int_{z}^{h}{E_{iGtB}d\; z^{\prime}}}} = {{\frac{\begin{matrix}{{\sin\left( {k_{iz}z} \right)} +} \\{r_{iB}{\cos\left( {k_{iz}z} \right)}}\end{matrix}}{h}{\int_{z}^{h}{z^{\prime}\;{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}}} - {r_{iT}z^{\prime}\;{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}d\; z^{\prime}}}} & (180)\end{matrix}$

Once more, using ID-8a and ID-9a in the UsefullIdentities documents:

$\begin{matrix}{\;{= {\frac{\begin{matrix}{{\sin\left( {k_{iz}z} \right)} +} \\{r_{iB}{\cos\left( {k_{iz}z} \right)}}\end{matrix}}{h}\left\lbrack {{{- \frac{z^{\prime}}{k_{iz}}}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} + {\frac{1}{k_{iz}^{2}}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} -} \right.}}} & {(181)} \\\left. {{\frac{r_{iT}z^{\prime}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} - {\frac{r_{iT}}{k_{iz}^{2}}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}} \right\rbrack_{z}^{h} & \\{= {\frac{\begin{matrix}{{\sin\left( {k_{iz}z} \right)} +} \\{r_{iB}{\cos\left( {k_{iz}z} \right)}}\end{matrix}}{k_{iz}h}\left( {{z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + \frac{1}{k_{iz}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} -} \right.}} & {(182)} \\\left. {{r_{iT}h} + {r_{iT}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right) & \end{matrix}$

Bringing the leading RHS denominator over to the LHS and thenmultiplying (182) out, term by term:

$\begin{matrix}{{\frac{k_{iz}h}{T_{iL}}{\int_{0}^{z}{E_{iGtB}{dz}^{\prime}}}} = {{z\;{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{1}{k_{iz}}{\sin\left( {k_{iz}z} \right)}} - {\frac{1}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}h{\sin\left( {k_{iz}z} \right)}} + {r_{iT}z{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}z{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}r_{iB}h{\cos\left( {k_{iz}z} \right)}} + {r_{iT}r_{iB}z{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (183)\end{matrix}$

We need to add the field due to current below z to the field due tocurrent above z. This means we must subtract (183) from (179), recallthe minus sign leading the LHS of (180). The first terms (for all butthe last two lines) come from (179) and the second terms come from(183). The next to last line comes from (179) and the last line comesfrom (183). For notational purposes, we also define Z₂ (z) in (184), forconvenience of notation.

$\begin{matrix}{\mspace{85mu}{{\frac{k_{iz}h}{T_{it}}\left( {{\int_{0}^{Z}{E_{iGtT}{dz}^{\prime}}} + {\int_{Z}^{h}{E_{iGtB}{dz}^{\prime}}}} \right)} = {Z_{2}(z)}}} & (184) \\{\mspace{85mu}{{\left( {{\int_{0}^{z}{E_{iGtT}{dz}^{\prime}}} + {\int_{z}^{h}{E_{iGtB}{dz}^{\prime}}}} \right) = {\frac{T_{it}}{k_{iz}h}{Z_{2}(z)}}}{{Z_{2}(z)} = {{z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {z{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{1}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} + {\frac{1}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {r_{iT}z{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}z{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iT}r_{iB}z{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iT}r_{iB}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {\frac{r_{iT}r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{1}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}}}}}} & (185) \\{\mspace{79mu}{{{- \frac{1}{k_{iz}}}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}h{\sin\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} + {r_{iT}r_{iB}h{\cos\left( {k_{iz}z} \right)}}}} & (186)\end{matrix}$

Noting that some terms cancel and we can apply ID-1 and ID-2 to otherterms:

$\begin{matrix}{{Z_{2}(z)} = {\frac{1}{k_{iz}}\left\{ {{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}} - {\sin\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {{r_{iT}k_{iz}h} - 1} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} \right\}}} & (187)\end{matrix}$

Using (185) we integrate the Green's function, (126) and (127), toobtain the transverse fields around a New Tapered Via. Then wesubstitute in (124) for T_(it):

$\begin{matrix}{\mspace{85mu}{E_{itTVIA} = {{\left( {{\int_{0}^{z}{E_{iGtT}{dz}^{\prime}}} + {\int_{z}^{h}{E_{iGtB}{dz}^{\prime}}}} \right)e_{i}} = {\frac{T_{it}}{k_{iz}h}{Z_{2}(z)}e_{i}}}}} & (188) \\{E_{tTV} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}h}\left( \frac{Z_{2}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right)e_{i}}}} & (189)\end{matrix}$

Substituting (187) for Z₂ (z), we have:

$\begin{matrix}{E_{tTV} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}h}\left( {1 - \frac{\begin{matrix}{{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}\cos\left( {k_{iz}\left( {h - z} \right)} \right)} +} \\{\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)e_{i}}}} & (190)\end{matrix}$

Given that a large portion of Z₂ (z) cancelled with the denominator, fornotational convenience going forward, we now define Z₃ (z), whichcontains all of the z variation of E

$\begin{matrix}{{Z_{3}(z)} = {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}}} & (191) \\{E_{tTV} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}h}\left( {1 - \frac{Z_{3}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)e_{i}}}} & (192)\end{matrix}$

Coupling Between the Tapered Via and Surface Rooftops

The coupling to an x-directed rooftop located at (x₁, y_(i), z) due to aTapered Via centered at (x₀, y₀) follows from (192):

$\begin{matrix}{{S_{TVtoRFX} = {\int{{{E_{tTV}(z)} \cdot J_{RFX}}{dxdy}}}}} & (193) \\{S_{TVtoRFX} = {\sum\limits_{TM}{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{2}{g_{1}\left( {x_{1},y_{1}} \right)}\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}h}\left( {1 - \frac{Z_{3}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)}}} & (194)\end{matrix}$

Similarly, coupling to ay-directed rooftop is (changes from (194) notedin yellow):-+

$\begin{matrix}{S_{TVtoRFY} = {\sum\limits_{TM}{{F_{R}\left( {\Delta x} \right)}{F_{T}\left( {\Delta y} \right)}N_{1}{g_{2}\left( {x_{1},y_{1}} \right)}\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}k_{iz}h}\left( {1 - \frac{Z_{3}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)}}} & (195)\end{matrix}$

For coupling to surface rooftops located at the bottom (zero current)end of the tapered via, we evaluate (189b), Z₃ (Z), at z=0:

Z ₃(0)=sin(k _(iz) h)+r _(iT) cos(k _(iz) h))+r _(iB) −r _(iT) r _(iB) k_(iz) h  (196)

For coupling to surface rooftops located at the top (maximum current)end of the tapered via, we evaluate (191), Z₃ (Z), at z=h:

Z ₃(h)=r _(IT)+(1−r _(iT) k _(iz) h)(sin(k _(iz) h)+r _(iB) cos(k _(iz)h))  (196a)

It is interesting to note that:

$\begin{matrix}{{\lim\limits_{h\rightarrow 0}{Z_{3}(0)}} = {{\lim\limits_{h\rightarrow 0}{Z_{3}(h)}} = {r_{iT} + r_{iB}}}} & \left( {196b} \right)\end{matrix}$

However, (194) and (195) both still go to zero as h goes to zero. It maybe of interest to explore

$\frac{1}{h}{\lim\limits_{h\rightarrow 0}{Z_{3}(z)}}$

if we wish to obtain non-zero matrix elements for zero substratethickness.

Coupling Between the Tapered Via and Volume Rooftops

For this problem, we must obtain the volume voltage by integrating thetransverse electric fields, (192), due to a tapered via over the lengthof the via. Thus we need to integrate Z₃ (z), (191), over z from 0 to h:

$\begin{matrix}{{\int_{0}^{h}{{Z_{3}(z)}{dz}}} = {\int_{0}^{h}{\left\{ {{\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} \right\}{dz}}}} & (197) \\{\mspace{85mu}{= {\frac{1}{k_{iz}}\begin{bmatrix}{{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}\sin\left( {k_{iz}\left( {h - z} \right)} \right)} +} \\{\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{- {\cos\left( {k_{iz}z} \right)}} + {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)}\end{bmatrix}}_{z = 0}^{h}}} & (198) \\{{k_{iz}{\int_{0}^{h}{{Z_{3}(z)}{dz}}}} = {1 - {\cos\left( {k_{iz}h} \right)} + {r_{iT}{\sin\left( {k_{iz}h} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)} + {r_{iB}{\sin\left( {k_{iz}h} \right)}}} \right)}}} & (199) \\{{k_{iz}{\int_{0}^{h}{{Z_{3}(z)}{dz}}}} = {{\left( {2 - {r_{iT}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} + {\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right){\sin\left( {k_{iz}h} \right)}}}} & (200)\end{matrix}$

Note that, in the limit as h goes to zero, as long as we divide by h, wehave the same results as for surface rooftops, (196b):

$\begin{matrix}{{\lim\limits_{h\rightarrow 0}{\frac{1}{h}{\int_{0}^{h}{{Z_{3}(z)}dz}}}} = {r_{iT} + r_{iB}}} & (201)\end{matrix}$

The coupling to an x-directed volume rooftop, VRFX, located at (x₁,y_(i)) due to a Tapered Via, TV, centered at (x₀, y₀) follows from (192)and (200). We must not forget to integrate the constant, 1, that leadsthe quantity in the large parentheses of (192) over the thickness of thesubstrate. The integration (202) is accomplished, as usual, by simplymultiplying by the Fourier coefficients:

$\begin{matrix}{{S_{TVIAtoVRFX} = {\int{{{E_{tTV}(z)} \cdot J_{RFX}}{dxdy}}}}} & (202) \\{S_{TVtoVRFX} = {\sum\limits_{TM}{\frac{{F_{T}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{2}{g_{1}\left( {x_{1},y_{1}} \right)}}{Y_{i}k_{iz}^{2}h}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0\prime}y_{0}} \right)}\left( {{k_{iz}h} - \frac{\begin{matrix}{{\left( {2 - {r_{iT}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} +} \\{\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right){\sin\left( {k_{iz}h} \right)}}\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)}}} & (203)\end{matrix}$

Similarly, coupling to a y-directed volume rooftop is (changes from(203) are highlighted):

$\begin{matrix}{S_{TVtoVRFY} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{T}\left( {\Delta y} \right)}N_{1}{g_{2}\left( {x_{1},y_{1}} \right)}}{Y_{i}k_{iz}^{2}h}{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}{g_{3}\left( {x_{0\prime}y_{0}} \right)}\left( {{k_{iz}h} - \frac{\begin{matrix}{{\left( {2 - {r_{iT}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} +} \\{\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right){\sin\left( {k_{iz}h} \right)}}\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} \right)}}} & (204)\end{matrix}$

Unfortunately, even if (203) and (204) are divided by the substratethickness, h, then they both still go to zero as h goes to zero, inspite of (201). It will likely be productive to explore dividing by thesquare of the substrate thickness.

This result is equivalent to (98) in the VolumeRooftops document. Inthat document, note that the k_(iz) ³ in the first denominator becomesof k_(iz) ² when it cancels with the k_(iz) in C_(i)(x₁, y₁), (90), thusmatching the result above.

Normal (z-directed) Fields of the Tapered Via

The current on the Tapered Via (TV) basis function has maximum currentat the top end of the via and tapers down to zero at the bottom end ofthe via, (175), repeated here:

$\begin{matrix}{J_{zTV} = \frac{z}{h}} & (205)\end{matrix}$

In a manner similar to the Tapered Via transverse field derivation,above, we must integrate (128) from z′=0 to z in order to get the fieldat a given z due to current below z, with (128) multiplied by (205).E_(iGzT) is the i^(th) term of the summation representing the Green'sfunction. For convenience, we let E_(iGzT)=E_(iGzT)u_(z,) and we workwith E_(iGzT)

$\begin{matrix}{{\frac{1}{T_{iz}}{\int_{0}^{z}{E_{iGzT}{dz}^{\prime}}}} = {{\frac{{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}{h}{\int_{0}^{z}{z^{\prime}{\cos\left( {k_{iz}z^{\prime}} \right)}}}} - {r_{iB}z^{\prime}{\sin\left( {k_{iz}z^{\prime}} \right)}{dz}^{\prime}}}} & (206)\end{matrix}$

Using identities ID-8a and ID-9a in the UsefullIdentities document:

$\begin{matrix}{{\frac{1}{T_{iz}}{\int_{0}^{z}{E_{iGzT}{dz}^{\prime}}}} = {{\frac{{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}{k_{iz}h}\left\lbrack {{z^{\prime}\sin\left( {k_{iz}z^{\prime}} \right)} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z^{\prime}} \right)}} + {r_{iB}z^{\prime}{\cos\left( {k_{iz}z^{\prime}} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z^{\prime}} \right)}}} \right\rbrack}\begin{matrix}\;^{z} & \; \\\;_{0} & \;\end{matrix}}} & (207) \\{{\frac{k_{iz}h}{T_{iz}}{\int_{0}^{z}{E_{iGzT}{dz}^{\prime}}}} = {\left( {{\cos\left( {k_{iz}\left( {h - z} \right)} \right)} - {r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)\left( {{{zsin}\left( {k_{iz}z} \right)} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} - \frac{1}{k_{iz}} + {r_{iB}z\;{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}}} \right)}} & (208)\end{matrix}$

Multiplying (208) out, term by term:

$\begin{matrix}{{\frac{k_{iz}h}{T_{iz}}{\int_{0}^{z}{E_{iGzT}{dz}^{\prime}}}} = {{z\;{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}{{z\cos}\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {r_{iT}z\;{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - \mspace{56mu}{r_{iT}r_{iB}z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} + {\frac{r_{iT}r_{iB}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}}}} & (209)\end{matrix}$

We must also calculate the field at z due to via current above z. Forthis, we integrate (130) from z′=z to h in order to get the field at zdue to current above z. As above, (130) is multiplied by (205) and wework with the magnitude of the i^(th) term.

$\begin{matrix}{{\frac{1}{T_{iz}}{\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} = {{\frac{{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}}{h}{\int_{z}^{h}{z^{\prime}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}}} - {r_{iT}z^{\prime}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}{dz}^{\prime}}}} & (210)\end{matrix}$

Using ID-8b and ID-9b in the UsefullIdentities documents:

$\begin{matrix}{\left. {{\frac{1}{T_{iz}}{\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} = {{\frac{{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}}{k_{iz}h}\left\lbrack {{{- z^{\prime}}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} + {\frac{1}{k_{iz}}\left( {h - z^{\prime}} \right)}} \right)} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} - {r_{iT}z^{\prime}{\cos\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z^{\prime}} \right)} \right)}}}} \right\rbrack\begin{matrix}\;^{h} & \; \\\;_{z} & \;\end{matrix}} & (211) \\{\left. {{\frac{k_{iz}h}{T_{iz}}{\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} = {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}}} \right)\left( {{z{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + \frac{1}{k_{iz}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}h} + {r_{iT}z{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} \right)} & (212)\end{matrix}$

Multiplying (212) out, term by term:

$\begin{matrix}{= {{z{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}h{\cos\left( {k_{iz}z} \right)}} + {r_{iT}z{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iB}z{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}} + {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iT}r_{iB}h{\sin\left( {k_{iz}z} \right)}} - {r_{iT}r_{iB}z{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iT}r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}} & (213)\end{matrix}$

We need to add the field due to current below z to the field due tocurrent above z. This means we must add (209) to (213). The first terms(for all but the last two lines) come from (209) and the second termscome from (213). The next to last line comes from (209) and the lastline comes from (213). For notational purposes, we also define Z₄(z) in(214), for convenience of notation.

$\begin{matrix}{{Z_{4}(z)} = {{\frac{k_{iz}h}{T_{iz}}\left( {{\int_{0}^{z}{E_{iGzT}{dz}^{\prime}}} + {\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} \right)} = {{{{zcos}\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {{{zcos}\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}{{zsin}\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}{{zcos}\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} + {\frac{r_{iT}}{k_{iz}}{\cos\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {r_{iB}{{zcos}\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}{{zsin}\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{r_{iB}}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} + {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} - {r_{iT}r_{iB}{{zsin}\left( {k_{iz}\left( {h - z} \right)} \right)}{\cos\left( {k_{iz}z} \right)}} - {r_{iT}r_{iB}{{zsin}\left( {k_{iz}z} \right)}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}r_{iB}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}{\sin\left( {k_{iz}z} \right)}} - {\frac{r_{iT}r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}}} & (214) \\{{{+ \frac{1}{k_{iz}}}{\cos\left( {k_{iz}z} \right)}} - {r_{iT}h{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}r_{iB}h{\sin\left( {k_{iz}z} \right)}}} & (215)\end{matrix}$

Noting that some terms cancel and others combine by applying ID-1 andID-2:

$\begin{matrix}{{Z_{4}(z)} = {{{z\left( {1 - {r_{iT}r_{iB}}} \right)}{\sin\left( {k_{iz}h} \right)}} + {{z\left( {r_{iT} + r_{iB}} \right)}{\cos\left( {k_{iz}h} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} - {r_{iT}h{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}} + {r_{iT}r_{iB}h{\sin\left( {k_{iz}z} \right)}}}} & (216)\end{matrix}$

For convenience in what follows, we reform (216) and define Z₅(z):

$\begin{matrix}{{Z_{4}(z)} = {{\frac{k_{iz}h}{T_{iz}}\left( {{\int_{0}^{z}{E_{iGzT}d{z'}}} + {\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} \right)} = {{z\left\{ {{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right\}} + {\frac{1}{k_{iz}}{Z_{5}(z)}}}}} & (217) \\{{Z_{5}(z)} = {{r_{iT}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\cos\left( {k_{iz}\left( {h - z} \right)} \right)} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right)}}} & (218)\end{matrix}$

The total normal field at a given level, z, is (129) added to the fielddue to via current from both below and above z. We start with (217):

$\begin{matrix}{\left( {{\int_{0}^{z}{E_{iGzT}d{z'}}} + {\int_{z}^{h}{E_{iGzB}{dz}^{\prime}}}} \right) = {{\frac{T_{iz}}{k_{iz}}\frac{z}{h}\left\{ {{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} \right\}} + {\frac{T_{iz}}{k_{iz}^{2}h}{Z_{5}(z)}}}} & (223)\end{matrix}$

Reforming the summation over all TM modes of the Green's function,combining (128)-(130) including (129) multiplied by (205), theintegration result of (217) gives us the z-directed field for the newuniform via:

$\begin{matrix}{E_{zTV} = {\sum\limits_{TM}{\left\{ {{T_{iz}\frac{z}{h}\frac{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}{k_{iz}}} + {T_{iz}\frac{Z_{5}(z)}{k_{iz}^{2}h}} - {T_{iv}\frac{z}{h}\left( {1 - \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right)}} \right\} u_{z}}}} & (224)\end{matrix}$

Substituting in (125a), which puts T_(iz) in terms of T_(iv):

$\begin{matrix}{E_{zTV} = {\sum\limits_{TM}{T_{iv}\left\{ {\frac{z}{h} + {\frac{1}{k_{iz}h}\frac{Z_{5}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}}} - {\frac{z}{h}\left( {1 - \frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}} \right)}} \right\} u_{z}}}} & (225)\end{matrix}$

Noting that two z/h terms cancel and substituting in (125b) for T_(iv):

$\begin{matrix}{E_{zTV} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta x} \right)}{F_{R}\left( {\Delta y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{Y_{i}h}\left( {\frac{Z_{5}(z)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} + {k_{iz}z\frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}}} \right)u_{z}}}} & (226)\end{matrix}$

Coupling Between the Tapered Via and the Uniform Via

For this, we must integrate (226), the normal electric field due to thetapered via, over the thickness of the substrate. We start byintegrating Z₅(z), given in (218):

$\begin{matrix}{{\int_{0}^{h}{{Z_{5}(z)}dz}} = {{{\int_{0}^{h}{r_{iT}^{(0)}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}} - {\cos\left( {k_{iz}\left( {h - z} \right)} \right)} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\cos\left( {k_{iz}z} \right)} - {r_{iB}{\sin\left( {k_{iz}z} \right)}}} \right){dz}}} = {{\frac{1}{k_{iz}}\left\lbrack {{r_{iT}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\sin\left( {k_{iz}\left( {h - z} \right)} \right)} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}z} \right)} + {r_{iB}{\cos\left( {k_{iz}z} \right)}}} \right)}} \right\rbrack}\begin{matrix}\;^{h} & \; \\\;_{0} & \;\end{matrix}}}} & (227) \\{{k_{iz}{\int_{0}^{h}{{Z_{5}(z)}{dz}}}} = {{r_{iT}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} - {\sin\left( {k_{iz}h} \right)} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{\sin\left( {k_{iz}h} \right)} - {r_{iB}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)}} \right)}}} & (228)\end{matrix}$

Gathering terms:

k _(iz)∫₀ ^(h) Z ₅(z)dz=(r _(iT) −r _(iB) +r _(iT) r _(iB) k _(iz)h)(1−cos(k _(iz) h))−r _(iT) k _(iz) h sin(k _(iz) h)  (229)

We will use (229) to integrate (226), but we also need to integrate thelast term of (226):

$\begin{matrix}{{\int_{0}^{h}{k_{iz}z\frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}dz}} = {\frac{k_{iz}h^{2}}{2}\frac{j\omega\mu k_{iz}Y_{i}}{k_{c}^{2}}}} & (230)\end{matrix}$

Now, using (229) and (230), we integrate (226):

$\begin{matrix}{{\int_{0}^{h}{{E_{zTV} \cdot u_{z}}{dz}}} = {\sum\limits_{T\; M}{\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{Y_{i}k_{iz}h}\left( {\frac{\begin{matrix}{{\left( {r_{iT} - r_{iB} + {r_{iT}r_{iB}k_{iz}h}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} -} \\{r_{iT}k_{iz}{\sin\left( {k_{iz}h} \right)}}\end{matrix}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} + {\frac{k_{iz}^{2}h^{2}}{2}\frac{j\;\omega\;\mu\; k_{iz}Y_{i}}{k_{c}^{2}}}} \right)}}} & (231)\end{matrix}$

Moving the k_(iz)h in the denominator of the first fraction to insidethe large parentheses, and integrating over x and y, we obtain thedesired coupling between a Tapered Via centered at (x₀, y₀) and aUniform Via centered at (x₁, y_(i)):

$\begin{matrix}{S_{TVtoUV} = {\sum\limits_{T\; M}{\frac{{F_{R}^{2}\left( {\Delta\; y} \right)}{F_{R}^{2}\left( {\Delta\; y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( {\frac{{\left( {\frac{r_{iT} - r_{iB}}{k_{iz}h} + {r_{iT}r_{iB}}} \right)\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} - {r_{iT}{\sin\left( {k_{iz}h} \right)}}}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iB} + r_{iT}} \right){\cos\left( {k_{iz}h} \right)}}} + {\frac{j\;\omega\;\mu\; k_{iz}^{2}Y_{i}}{2k_{c}^{2}}h}} \right)}}} & (232)\end{matrix}$

Self-Coupling of the Tapered Via

For this, we must multiply (226), the normal electric field due to thetapered via, by z/h (from (175), (205)) and then integrate over thethickness of the substrate. Starting by integrating the first terminside the curly braces of (226), we must integrate z/h times Z₅ (z),(218). Integrating each term and using ID-8a,b and ID-9a,b in theUsefulIdenties document:

$\begin{matrix}{{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}} = {{\frac{1}{h}{\int_{0}^{h}{r_{iT}z\mspace{11mu}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}}}} - {z\;{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{z\mspace{11mu}{\cos\left( {k_{iz}z} \right)}} - {r_{iB}z\;{\sin\left( {k_{iz}z} \right)}}} \right){dz}}}} & (233) \\{{dz} = {\frac{1}{k_{iz}h}\left\lbrack {{r_{iT}z\;\cos\;\left( {k_{iz}\left( {h - z} \right)} \right)} + {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} + {z\;{\sin\left( {k_{iz}\left( {h - z} \right)} \right)}} - {\frac{1}{k_{iz}}{\cos\left( {k_{iz}\left( {h - z} \right)} \right)}} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{z\;{\sin\left( {k_{iz}z} \right)}} + {\frac{1}{k_{iz}}{\cos\left( {k_{iz}z} \right)}} + {r_{iB}z\mspace{11mu}{\cos\left( {k_{iz}z} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}z} \right)}}} \right\rbrack_{0}^{h}}} \right.}} & (234) \\{{k_{iz}h{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}}} = {{r_{iT}h} - {\frac{r_{iT}}{k_{iz}}{\sin\left( {k_{iz}h} \right)}} - {\frac{1}{k_{iz}}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} + {\left( {1 - {r_{iT}k_{iz}h}} \right)\left( {{h\;{\sin\left( {k_{iz}h} \right)}} - {\frac{1}{k_{iz}}\left( {1 - {\cos\left( {k_{iz}h} \right)}} \right)} + {r_{iB}h\;{\cos\left( {k_{iz}h} \right)}} - {\frac{r_{iB}}{k_{iz}}{\sin\left( {k_{iz}h} \right)}}} \right)}}} & (235)\end{matrix}$

Gathering terms and simplifying:

$\begin{matrix}{{k_{iz}h{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}}} = {{2r_{iT}h} - \frac{2}{k_{iz}} + {\left( {{- \frac{r_{iT}}{k_{iz}}} + h - {r_{iT}k_{iz}h^{2}} - \frac{r_{iB}}{k_{iz}} + {r_{iT}r_{iB}h}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {\frac{2}{k_{iz}} - {r_{iT}h} + {r_{iB}h} - {r_{iT}r_{iB}k_{iz}h^{2}}} \right){\cos\left( {k_{iz}h} \right)}}}} & (236) \\{= {{\frac{2}{k_{iz}}\left( {{r_{iT}k_{iz}h} - 1} \right)} + {\frac{1}{k_{iz}}\left( {{- r_{iT}} + {k_{iz}h} - {r_{iT}k_{iz}^{2}h^{2}} - r_{iB} + {r_{iT}r_{iB}k_{iz}h}} \right){\sin\left( {k_{iz}h} \right)}} + {\frac{1}{k_{iz}}\left( {2 - {r_{iT}k_{iz}h} + {r_{iB}k_{iz}h} - {r_{iT}r_{iB}k_{iz}^{2}h^{2}}} \right){\cos\left( {k_{iz}h} \right)}}}} & (237) \\{{k_{iz}^{2}h{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}}} = {{2\left( {{r_{iT}k_{iz}h} - 1} \right)} + {\left\{ {{\left( {{r_{iT}r_{iB}} + 1} \right)k_{iz}h} - {r_{iT}k_{iz}^{2}h^{2}} - r_{iT} - r_{iB}} \right\}{\sin\left( {k_{iz}h} \right)}} + {\left\{ {2 - {\left( {r_{iT} - r_{iB}} \right)k_{iz}h} - {r_{iT}r_{iB}k_{iz}^{2}h^{2}}} \right\}{\cos\left( {k_{iz}h} \right)}}}} & (238)\end{matrix}$

For notational convenience, we define:

$\begin{matrix}{{Z_{6}(h)} = {{\frac{1}{h}{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}}} = {\frac{1}{k_{iz}^{2}h^{2}}\left\lbrack {{2\left( {{r_{iT}k_{iz}h} - 1} \right)} + {\left\{ {{\left( {{r_{iT}r_{iB}} + 1} \right)k_{iz}h} - {r_{iT}k_{iz}^{2}h^{2}} - r_{iT} - r_{iB}} \right\}{\sin\left( {k_{iz}h} \right)}} + {\left\{ {2 - {\left( {r_{iT} - r_{iB}} \right)k_{iz}h} - {r_{iT}r_{iB}k_{iz}^{2}h^{2}}} \right\}{\cos\left( {k_{iz}h} \right)}}} \right\rbrack}}} & (239) \\{= {\frac{1}{k_{iz}^{2}h^{2}}\left\lbrack {{\left( {r_{iT} + r_{iB} - {r_{iT}r_{iB}k_{iz}h}} \right)k_{iz}h} + {\left\{ {{\left( {{r_{iT}r_{iB}} + 1} \right)k_{iz}h} - r_{iT} - r_{iB}} \right\}{\sin\left( {k_{iz}h} \right)}} + {\left\{ {2 - {\left( {r_{iT} - r_{iB}} \right)k_{iz}h} - {r_{iT}r_{iB}k_{iz}^{2}h^{2}}} \right\}\left( {{\cos\left( {k_{iz}h} \right)} - 1} \right)}} \right\rbrack}} & (240)\end{matrix}$

Note that defining Z₆(h) with the leading 1/h makes Z₆(h) dimensionless.The only difference between the above two equations is in the firstconstant in the expression inside the square brackets and thecos(k_(iz)h) factor being changed to (cos(k_(iz)h)−1), which arehighlighted in yellow.

We must also integrate the second term in the large parentheses of(226), after multiplying it by z/h:

$\begin{matrix}{{\int_{0}^{h}{\frac{k_{iz}z^{2}}{h}\frac{j\;\omega\;\mu\; j_{iz}Y_{i}}{k_{c}^{2}}{dz}}} = {{\frac{k_{iz}h^{2}}{3}\frac{j\;\omega\;\mu\; k_{iz}Y_{i}}{k_{c}^{2}}} = {\frac{j\;\omega\;\mu\; k_{iz}^{2}Y_{i}}{3k_{c}^{2}}h^{2}}}} & (241)\end{matrix}$

Integrating the magnitude of the i^(th) term of (226) multiplied by z/h,we have:

$\begin{matrix}{{\int_{0}^{h}{\frac{z}{h}E_{izTV}{dz}}} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{Y_{i}h}\begin{pmatrix}{\frac{h\left( {\frac{1}{h}{\int_{0}^{h}{\frac{z}{h}{Z_{5}(z)}{dz}}}} \right)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} +} \\{\int_{0}^{h}{\frac{k_{iz}z^{2}}{h}\frac{j\;\omega\;\mu\; k_{iz}Y_{i}}{k_{c}^{2}}{dz}}}\end{pmatrix}}} & (242)\end{matrix}$

Substituting in either (239) or (240), and (241), and also bringing theh that is in denominator of the leading fraction inside the largeparentheses, we have:

$\begin{matrix}{{\int_{0}^{h}{\frac{z}{h}E_{izTV}{dz}}} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x,y} \right)}}{Y_{i}}\left( {\frac{Z_{6}h}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} + {\frac{j\;\omega\;\mu\; k_{iz}^{2}Y_{i}}{3k_{c}^{2}}h}} \right)}} & (243)\end{matrix}$

To obtain the self-coupling matrix element values, we perform thedesired integration over the horizontal surface of the substrate bymultiplying by the usual Fourier coefficients:

$\begin{matrix}{S_{TVtoTV} = {\sum\limits_{TM}{\frac{{F_{R}^{2}\left( {\Delta\; x} \right)}{F_{R}^{2}\left( {\Delta\; y} \right)}N_{3}^{2}{g_{3}\left( {x_{0},y_{0}} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}{Y_{i}}\left( {\frac{Z_{6}(h)}{{\left( {1 - {r_{iT}r_{iB}}} \right){\sin\left( {k_{iz}h} \right)}} + {\left( {r_{iT} + r_{iB}} \right){\cos\left( {k_{iz}h} \right)}}} + {\frac{j\;\omega\;\mu\; k_{iz}^{2}Y_{i}}{3k_{c}^{2}}h}} \right)}}} & (244)\end{matrix}$

Appendix 2

Subsection Constants

Useful Identities

$\begin{matrix}{{\sin\left( {A + B} \right)} = {{{\sin(A)}{\cos(B)}} + {{\cos(A)}{\sin(B)}}}} & {{ID}\text{-}1} \\{{\cos\left( {A + B} \right)} = {{{\cos(A)}{\cos(B)}} - {{\sin(A)}{\sin(B)}}}} & {{ID}\text{-}2} \\{{\tan\left( \frac{A}{2} \right)} = \frac{1 - {\cos(A)}}{\sin(A)}} & {{ID}\text{-}3} \\{{\tan\left( \frac{A}{2} \right)} = \frac{\sin(A)}{1 + {\cos(A)}}} & {{ID}\text{-}4} \\{{{\sin(A)}{\sin(B)}} = {\frac{1}{2}\left\lbrack {{\cos\left( {A - B} \right)} - {\cos\left( {A + B} \right)}} \right\rbrack}} & {{ID}\text{-}5} \\{{{\cos(A)}{\cos(B)}} = {\frac{1}{2}\left\lbrack {{\cos\left( {A - B} \right)} + {\cos\left( {A + B} \right)}} \right\rbrack}} & {{ID}\text{-}6} \\{{{\sin(A)}{\cos(B)}} = {\frac{1}{2}\left\lbrack {{\sin\left( {A + B} \right)} + {\sin\left( {A - B} \right)}} \right\rbrack}} & {{ID}\text{-}7} \\{{{\cos(A)}{\sin(B)}} = {\frac{1}{2}\left\lbrack {{\sin\left( {A + B} \right)} - {\sin\left( {A - B} \right)}} \right\rbrack}} & {{ID}\text{-}7a} \\{{\int{{\sin\left( {ax} \right)}dx}} = {{{- \frac{1}{a}}{\cos\left( {ax} \right)}} + C}} & {{ID}\text{-}8} \\{{\int{x{\sin\left( {ax} \right)}dx}} = {{{- \frac{x}{a}}{\cos\left( {ax} \right)}} + {\frac{1}{a^{2}}{\sin\left( {ax} \right)}} + C}} & {{ID}\text{-}8a} \\{{\int{{\cos\left( {ax} \right)}dx}} = {{\frac{1}{a}{\sin\left( {ax} \right)}} + C}} & {{ID}\text{-}9} \\{{\int{x{\cos\left( {ax} \right)}dx}} = {{\frac{x}{a}{\sin\left( {ax} \right)}} + {\frac{1}{a^{2}}{\cos\left( {ax} \right)}} + C}} & {{ID}\text{-}9a} \\{{\nabla{\times A}} = \left| \begin{matrix}u_{\chi} & u_{y} & u_{Z} \\{\partial{/{\partial x}}} & {\partial{/{\partial y}}} & {\partial{/{\partial z}}} \\A_{x} & A_{y} & A_{z}\end{matrix} \right|} & {{ID}\text{-}10} \\{{\nabla{\times H}} = {{j\;\omega\;\epsilon\; E} + J}} & {{ID}\text{-}11} \\{{\nabla{\times E}} = {{- j}\;{\omega\mu}\; H}} & {{ID}\text{-}12}\end{matrix}$

Basic Variables

Index i is used for summing over all m, n, TE, TM modes.

Index j goes over all dielectric layers, starting with layer j=0 at thetop and increases going down. The z coordinate starts at zero at thebottom and increases going up.

Wavenumber k=2π/λ=ω√{square root over (με)} in general. Modaladmittances are the ratio of H over E of the standing wave modes, andthus differ by a factor of j from the usual traveling wave modaladmittances.

$\begin{matrix}{k_{z,i}^{(j)} = \sqrt{k_{j}^{2} - k_{c}^{2}}} & (1) \\{k_{c}^{2} = {{\left( \frac{m\pi}{a} \right)^{2} + \left( \frac{n\pi}{b} \right)^{2}} = {k_{x}^{2} + k_{y}^{2}}}} & \left( {1a} \right) \\{y_{i,{TE}}^{(j)} = \frac{k_{z,i}^{(j)}}{j\omega\mu_{j}}} & (2) \\{Y_{i,{TM}}^{(j)} = {- \frac{j\;{\omega\epsilon}_{j}}{k_{z,i}^{(j)}}}} & (3)\end{matrix}$

Transverse normalized modal vectors:

$\begin{matrix}{\mspace{85mu}{{e_{i,{TE}} = {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{y}y} \right)}u_{x}\mspace{14mu} m} = 0}},{n > 0},}} & (4) \\{\mspace{85mu}{{= {{{- \sqrt{\frac{2}{ab}}}{\sin\left( {k_{x}x} \right)}u_{y}\mspace{20mu} m} > 0}},{n = 0},\mspace{20mu}{{{otherwise}\mspace{14mu}{for}\mspace{14mu} m\mspace{14mu}{and}\mspace{14mu} n} > {0\text{:}}}}} & (5) \\{= {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{\frac{n}{b}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{x}} - {\frac{m}{a}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (6) \\{\mspace{79mu}{{h_{i,{TE}} = {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{y}y} \right)}u_{y}\mspace{14mu} m} = 0}},{n > 0},}} & (7) \\{\mspace{79mu}{{= {{\sqrt{\frac{2}{ab}}{\sin\left( {k_{x}x} \right)}u_{x}\mspace{20mu} m} > 0}},{n = 0},\mspace{20mu}{{{otherwise}\mspace{14mu}{for}\mspace{14mu} m\mspace{14mu}{and}\mspace{14mu} n} > {0\text{:}}}}} & (8) \\{= {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{\frac{m}{a}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{x}} + {\frac{n}{b}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (9)\end{matrix}$

For the TM modal vectors, both m and n>0:

$\begin{matrix}{e_{i,{TM}} = {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{\frac{m}{a}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{x}} + {\frac{n}{b}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (10) \\{\mspace{85mu}{= {{N_{2}g_{1}u_{x}} + {N_{1}g_{2}u_{y}}}}} & \left( {10a} \right) \\{h_{i,{TM}} = {2{{\sqrt{\frac{ab}{{n^{2}a^{2}} + {m^{2}b^{2}}}}\left\lbrack {{{- \frac{n}{b}}{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}y} \right)}u_{x}} + {\frac{m}{a}{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}y} \right)}u_{y}}} \right\rbrack}.}}} & (11) \\{\mspace{85mu}{= {{{- N_{1}}g_{2}u_{x}} + {N_{2}g_{1}u_{y}}}}} & \left( {11a} \right) \\{{{{where}\mspace{14mu} N_{1}} = \frac{2k_{y}}{k_{c}\sqrt{ab}}},{N_{2} = \frac{2k_{x}}{k_{c}\sqrt{ab}}},{g_{1} = {{\cos\left( {k_{x}x} \right)}{\sin\left( {k_{y}x} \right)}}},{{{and}\mspace{14mu} g_{2}} = {{\sin\left( {k_{x}x} \right)}{\cos\left( {k_{y}x} \right)}}}} & \left( {11b} \right)\end{matrix}$

Transverse fields for the j^(th) layer and the i^(th) (m, n, TE, TM)mode:

E _(t,i) ^((j)) ={F _(i) ^((j))sin(k _(z,i) ^((j)) z)+G _(i) ^((j))cos(k_(z,i) ^((j)) z)}e _(i),  (12)

H _(t,i) ^((j)) =−Y _(i) ^((j)) {F _(i) ^((j))cos(k _(z,i) ^((j)))−G_(i) ^((j))sin(k _(z,i) ^((j)) z}h _(i).  (13)

F and G are set to match boundary conditions at top and bottom covers.If we change z to (c−z), then also change sign of H.

Important Integrals and Functions

Rectangle Function:

$\begin{matrix}{{R\left( {x,x_{0},{\Delta\; x}} \right)} = \begin{matrix}{1,} & {{x_{0} - {\Delta\; x}} < x < {x_{0} + {\Delta\; x}}} \\{0,} & {Otherwise}\end{matrix}} & (14)\end{matrix}$

Sine Rectangle Integral:

$\begin{matrix}{{S_{R}\left( {x_{0},{\Delta\; x}} \right)} = {\int_{x = {x_{0} - {\Delta\; x}}}^{x = {x_{0} + {\Delta\; x}}}{{\sin\left( {k_{x}x} \right)}{dx}}}} & (15) \\{= {{- {\frac{1}{k_{x}}\left\lbrack {\cos\left( {k_{x}x} \right)} \right\rbrack}_{x = {x_{0} - {\Delta\; x}}}^{x = {x_{0} + {\Delta\; x}}}}\mspace{14mu}\left( {{by}\mspace{14mu}{ID}\text{-}8} \right)}} & (16) \\{{= {{- \frac{1}{k_{x}}}\left( {{\cos\left( {k_{x}\left( {x_{0} + {\Delta\; x}} \right)} \right)} - {\cos\left( {k_{x}\left( {x_{0} - {\Delta\; x}} \right)} \right)}} \right)}}\left( {{next},{{apply}\mspace{14mu}{ID}\text{-}2\mspace{14mu}{twice}}} \right)} & (17) \\{= {\frac{2}{k_{x}}{\sin\left( {k_{x}\Delta\; x} \right)}{\sin\left( {k_{x}x_{0}} \right)}}} & (18)\end{matrix}$

Cosine Rectangle Integral:

$\begin{matrix}{{C_{R}\left( {x_{0},{\Delta\; x}} \right)} = {\int_{x = {x_{0} - {\Delta x}}}^{x = {x_{0} + {\Delta x}}}{{\cos\left( {k_{x}x} \right)}{dx}}}} & {(19)} \\{= {\frac{1}{k_{x}}\left\lbrack {\sin\left( {k_{x}x} \right)} \right\rbrack}_{x = {x_{0} - {\Delta x}}}^{x = {x_{0} + {\Delta x}}}} & {(20)} \\{\left( {{by}\mspace{14mu}{ID}\text{-}9} \right)} & \\{= {\frac{1}{k_{x}}\left( {{\sin\left( {k_{x}\left( {x_{0} + {\Delta x}} \right)} \right)} - {\sin\left( {k_{x}\left( {x_{0} - {\Delta x}} \right)} \right)}} \right)}} & {(21)} \\{\left( {{next},{{apply}\mspace{20mu}{ID}\text{-}7a}} \right)} & \\{= {\frac{2}{k_{x}}{\sin\left( {k_{x}\Delta x} \right)}{\cos\left( {k_{x}x_{0}} \right)}}} & {{~~~~~~~~~~~~~~~~~~~~}(23)}\end{matrix}$

Rectangle Fourier Coefficient:

$\begin{matrix}{{F_{R}\left( {\Delta x} \right)} = \begin{matrix}{{\frac{2}{k_{x}}\sin\;\left( {k_{x}\Delta\; x} \right)},} & {k_{x} \neq 0} \\{{\Delta\; x},} & {k_{x} = 0}\end{matrix}} & (24)\end{matrix}$

Triangle Function:

$\begin{matrix}{\ {{T\left( {x,x_{0},{\Delta\; x}} \right)} = \begin{matrix}{{\frac{x}{2\Delta x} + \left( {1 - \frac{x_{0}}{2\Delta x}} \right)}\ ,} & {{x_{0} - {2\Delta x}} < x < x_{0}} \\{{{- \frac{x}{2\Delta x}} + \left( {1 + \frac{x_{0}}{2\Delta x}} \right)},} & {x_{0} < x < {x_{0} + {2\Delta\; x}}} \\{0,} & {Otherwise}\end{matrix}}} & (25)\end{matrix}$

Sine Triangle Integral (note, all cos terms cancel in (28)):

$\begin{matrix}{{S_{T}\left( {x_{0},{\Delta\; x}} \right)}=={{\int_{x_{0} - {2{\Delta x}}}^{x = x_{0}}{\frac{x}{2\Delta\; x}\ {\sin\left( {k_{x}x} \right)}{dx}}} + {\int_{x_{0} - {2{\Delta x}}}^{x = x_{0}}{\left( {1 - \frac{x_{0}}{2\Delta x}} \right){\sin\left( {k_{x}x} \right)}{dx}}} + {\int_{x = x_{0}}^{x_{0} + {2{\Delta x}}}{\left( {{- \ \frac{x}{2\Delta x}}{\sin\left( {k_{x}x} \right)}} \right){dx}}} +}} & {{~~}(26)} \\{\int_{x = x_{0}}^{x_{0} + {2{\Delta x}}}{\left( {1 + \frac{x_{0}}{2\Delta x}} \right){\sin\left( {k_{x}x} \right)}{dx}}} & \\{= {{\frac{1}{2\Delta\;{xk}_{x}}\left\lbrack {{{- x}\;{\cos\left( {k_{x}x} \right)}} + {\frac{1}{k_{x}}{\sin\left( {k_{x}x} \right)}} - {\left( {{2\Delta\; x} - x_{0}} \right){\cos\left( {k_{x}x} \right)}}} \right\rbrack}_{x = {x_{0} - {2\Delta\; x}}}^{x = x_{0}} + {\frac{1}{2\Delta\;{xk}_{x}}\left\lbrack {{x\;{\cos\left( {k_{x}x} \right)}} - \frac{1}{k_{x}}} \right.}}} & {(27)} \\\left. {{\sin\left( {k_{x}x} \right)} - {\left( {{2\Delta\; x} + x_{0}} \right){\cos\left( {k_{x}x} \right)}}} \right\rbrack_{x = x_{0}}^{x = {x_{0} + {2\Delta\; x}}} & \; \\{= {\frac{1}{2\Delta\;{xk}_{x}}\begin{Bmatrix}{{{- x_{0}}{\cos\left( {k_{x}x_{0}} \right)}} + {\frac{1}{k_{x}}{\sin\left( {k_{x}x_{0}} \right)}} - {\left( {{2\Delta\; x} - x_{0}} \right){\cos\left( {k_{x}x_{0}} \right)}} + \ldots +} \\{{\left( {x_{0} - {2\Delta\; x}} \right){\cos\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} - {\frac{1}{k_{x}}{\sin\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} + {\left( {{2\Delta\; x} - x_{0}} \right){\cos\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} + \ldots +} \\{{\left( {x_{0} + {2\Delta\; x}} \right){\cos\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} - {\frac{1}{k_{x}}{\sin\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} - {\left( {{2\Delta\; x} + x_{0}} \right){\cos\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} + \ldots +} \\{{{- x_{0}}{\cos\left( {k_{x}x_{0}} \right)}} + {\frac{1}{k_{x}}{\sin\left( {k_{x}x_{0}} \right)}} + {\left( {{2\Delta\; x} + x_{0}} \right){\cos\left( {k_{x}x_{0}} \right)}}}\end{Bmatrix}}} & {{~~}(28)} \\{= {\frac{1}{\Delta\;{xk}_{x}^{2}}\left( {1 - {\cos\left( {k_{x}\Delta\; x} \right)}} \right){\sin\left( {k_{x}x_{0}} \right)}}} & {{~~~~~~~~~~}(29)}\end{matrix}$

Cosine Triangle Integral (note, all sin terms cancel in (32)):

$\begin{matrix}{{C_{T}\left( {x_{0},{\Delta\; x}} \right)}=={{\int_{x_{0} - {2{\Delta x}}}^{x = x_{0}}{\frac{x}{2\Delta\; x}{\cos\left( {k_{x}x} \right)}{dx}}} + {\int_{x_{0} - {2{\Delta x}}}^{x = x_{0}}{\left( {1 - \frac{x_{0}}{2\Delta x}} \right){\cos\left( {k_{x}x} \right)}{dx}}} + {\int_{x = x_{0}}^{x_{0} + {2{\Delta x}}}{\left( {{- \ \frac{x}{2\Delta x}}{\cos\left( {k_{x}x} \right)}} \right){dx}}} +}} & {(30)} \\{\int_{x = x_{0}}^{x_{0} + {2{\Delta x}}}{\left( {1 + \frac{x_{0}}{2\Delta x}} \right){\cos\left( {k_{x}x} \right)}{dx}}} & \\{= {{\frac{1}{2\Delta\;{xk}_{x}}\left\lbrack {{x\;{\sin\left( {k_{x}x} \right)}} + {\frac{1}{k_{x}}{\cos\left( {k_{x}x} \right)}} + {\left( {{2\Delta\; x} - x_{0}} \right){\sin\left( {k_{x}x} \right)}}} \right\rbrack}_{x = {x_{0} - {2\Delta\; x}}}^{x = x_{0}} + {\frac{1}{2\Delta\;{xk}_{x}}\left\lbrack {{{- x}\;{\sin\left( {k_{x}x} \right)}} - \frac{1}{k_{x}}} \right.}}} & {(31)} \\\left. {{\cos\left( {k_{x}x} \right)} + {\left( {{2\Delta\; x} + x_{0}} \right){\sin\left( {k_{x}x} \right)}}} \right\rbrack_{x = x_{0}}^{x = {x_{0} + {2\Delta\; x}}} & \\{= {\frac{1}{2\Delta\;{xk}_{x}}\begin{Bmatrix}{{x_{0}{\sin\left( {k_{x}x_{0}} \right)}} + {\frac{1}{k_{x}}{\cos\left( {k_{x}x_{0}} \right)}} + {\left( {{2\Delta\; x} - x_{0}} \right){\sin\left( {k_{x}x_{0}} \right)}} + \ldots -} \\{{\left( {x_{0} - {2\Delta\; x}} \right){\sin\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} - {\frac{1}{k_{x}}{\cos\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} - {\left( {{2\Delta\; x} - x_{0}} \right){\sin\left( {k_{x}\left( {x_{0} - {2\Delta\; x}} \right)} \right)}} + \ldots -} \\{{\left( {x_{0} + {2\Delta\; x}} \right){\sin\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} - {\frac{1}{k_{x}}{\cos\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} + {\left( {{2\Delta\; x} + x_{0}} \right){\sin\left( {k_{x}\left( {x_{0} + {2\Delta\; x}} \right)} \right)}} + \ldots +} \\{{x_{0}{\sin\left( {k_{x}x_{0}} \right)}} + {\frac{1}{k_{x}}{\cos\left( {k_{x}x_{0}} \right)}} - {\left( {{2\Delta\; x} + x_{0}} \right){\sin\left( {k_{x}x_{0}} \right)}}}\end{Bmatrix}}} & {(32)} \\{= {\frac{1}{\Delta\;{xk}_{x}^{2}}\left( {1 - {\cos\left( {k_{x}\Delta\; x} \right)}} \right){\cos\left( {k_{x}x_{0}} \right)}}} & {(33)}\end{matrix}$

Triangle Fourier Coefficient:

$\begin{matrix}{{F_{T}\left( {\Delta x} \right)} = \begin{matrix}{{\frac{1}{\Delta xk_{x}^{2}}\left( {1 - {\cos\left( {k_{x}\Delta x} \right)}} \right)},} & {k_{x} \neq 0} \\{{2\Delta\; x},} & {k_{x} = 0}\end{matrix}} & (34)\end{matrix}$

Appendix 3

Appendix 3: Multi-layer Coupling

For an overview of the field description used below, see the firstsection of the Appendix 1.

We assume a Source Subsection (i.e., a subsection with current on it) inlayer 1, which has a thickness of h₁. Given the fields in Layer 1, wedetermine the fields in Layer 0. Layer 0 contains the Upper FieldSubsection, the subsection to which we wish to determine the couplingfrom the source subsection. After completing that, we also determine thecoupling to the Lower Field Subsection, in Layer 2. See the Figurebelow.

To determine the fields in Layer 0 due to the Source Subsection, we needonly the transverse electric field due to the source at the surfacez=c₁. This is assumed to be known (as specified in Appendix 1) and ofthe form:

$\begin{matrix}{\left\lbrack E_{t}^{(1)} \right\rbrack_{z = c_{1}} = {\sum\limits_{i}{V_{iU}^{(1)}e_{i}}}} & (1)\end{matrix}$

The transverse fields for the top layer are:

$\begin{matrix}{E_{t}^{(0)} = {\sum\limits_{i}{V_{i}^{(0)}\left\{ {{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\} e_{i}}}} & (2) \\{H_{t}^{(0)} = {\sum\limits_{i}{V_{i}^{(0)}Y_{i}^{(0)}\left\{ {{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\} h_{i}}}} & (3)\end{matrix}$

The top cover surface impedance, i.e., the ratio of electric to magneticfield at the top cover, Level 0, z=c₀, is assumed to be known. The topcover impedance is normalized to the waveguide modal impedance as r_(iT)⁽⁰⁾=Y_(i) ⁽⁰⁾R_(iT) ⁽⁰⁾. Note that r_(iT) ⁰ is indeed the result if wetake the ratio of (2) over (3), multiply by the modal impedance andevaluate at z=c₀. The r_(iT) ⁽⁰⁾ impedance then determines all the lower(higher index) r_(iT) ^((j)) boundary impedances, which are r_(iT) ⁽⁰⁾transformed by the rectangular waveguide transmission line formed by thesidewalls. For example, r_(iT) ⁽¹⁾ is determined by taking the ratio of(2) over (3), multiplying by Y_(i) ⁽¹⁾ and evaluating at z=c₁ yielding:

$\begin{matrix}{r_{iT}^{(1)} = {{Y_{i}^{(1)}R_{iT}^{(1)}} = {\frac{Y_{i}^{(1)}}{Y_{i}^{(0)}}\frac{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{{\cos\left( {k_{iz}^{(0)}h_{0}} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}^{({j - 1})}h_{0}} \right)}}}}}} & \left( {4a} \right)\end{matrix}$

Lower (higher index) surface impedances are determined recursively as:

$\begin{matrix}{r_{iT}^{(j)} = {{Y_{i}^{(j)}R_{iT}^{(j)}} = {\frac{Y_{i}^{(j)}}{Y_{i}^{({j - 1})}}\frac{{\sin\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)} + {r_{iT}^{({j - 1})}{\cos\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)}}}{{\cos\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)} - {r_{iT}^{({j - 1})}{\sin\left( {k_{iz}^{({j - 1})}h_{j - 1}} \right)}}}}}} & \left( {4b} \right)\end{matrix}$

The analogous impedance of the bottom cover transformed up through thedielectric stack is also determined recursively:

$\begin{matrix}{r_{iB}^{(j)} = {{Y_{i}^{(j)}R_{iB}^{(j)}} = {\frac{Y_{i}^{(j)}}{Y_{i}^{({j + 1})}}\frac{{\sin\left( {k_{iz}^{({j + 1})}h_{j + 1}} \right)} + {r_{iB}^{({j + 1})}{\cos\left( {k_{iz}^{({j + 1})}h_{j + 1}} \right)}}}{{\cos\left( {k_{iz}^{({j + 1})}h_{j + 1}} \right)} - {r_{iB}^{({j + 1})}{\sin\left( {k_{iz}^{({j + 1})}h_{j + 1}} \right)}}}}}} & (5)\end{matrix}$

Our task is to determine the V_(i) ⁽⁰⁾ in terms of the known V_(iU) ⁽¹⁾.Evaluating the fields in both regions using (1) and (2) at Level 1 andsetting them equal:

$\begin{matrix}{\left\lbrack E_{t}^{(0)} \right\rbrack_{z = c_{1}} = \left\lbrack E_{t}^{(1)} \right\rbrack_{z = c_{1}}} & (6) \\{{V_{i}^{(0)}\left\{ {{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}} \right\}} = V_{iU}^{(1)}} & (7) \\{V_{i}^{(0)} = \frac{V_{iU}^{(1)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}} & (8)\end{matrix}$

Substituting (8) into (2) we have the transverse fields. We also listthe corresponding z-directed field, from (14) and (16) in Appendix 1:

$\begin{matrix}{E_{t}^{(0)} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}}{\begin{matrix}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} +} \\{r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}\end{matrix}}\left\{ {{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\} e_{i}}}} & (9) \\{E_{z}^{(0)} = {\sum\limits_{TM}{\frac{V_{iU}^{(1)}k_{iz}^{(0)}N_{3}}{\begin{matrix}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} +} \\{r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}\end{matrix}}\left\{ {{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\}{g_{3}\left( {x,y} \right)}u_{z}}}} & (10)\end{matrix}$

Since we have a c₀−z dependence, (10) is of opposite sign as compared to(14) in Appendix 1, as noted there. Analogously, we can find the fieldsin Layer 2, below the Source Layer. This layer contains the Lower FieldSubsection. All we need is the transverse fields from the SourceSubsection evaluated at Level 2, the bottom side of the Source Layer, inorder to determine all three field components in Layer 2:

$\begin{matrix}{\mspace{79mu}{\left\lbrack E_{t}^{(1)} \right\rbrack_{z = c_{2}} = {\sum\limits_{i}{V_{iL}^{(1)}e_{i}}}}} & (11) \\{E_{t}^{(2)} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}}{\begin{matrix}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} +} \\{r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}\end{matrix}}\left\{ {{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\} e_{i}}}} & (12) \\{E_{z}^{(2)} = {- {\sum\limits_{TM}{\frac{V_{iL}^{(1)}k_{iz}^{(2)}N_{3}}{\begin{matrix}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} +} \\{r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}\end{matrix}}\left\{ {{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{g_{3}\left( {x,y} \right)}u_{z}}}}} & (13)\end{matrix}$

Summations over i indicate summation over all TE and TM rectangularwaveguide modes. Note that the TE modes have no z-directed electricfield, so (10) and (13) are summed only over the TM modes. Thissituation arises only when we must deal with a via (Tapered Via or TV,or a Uniform Via, UV) as either the field or source subsection.

Note that in this document, (x₀, y₀) is always the center of the sourcesubsection and y_(i)) is the center of the field subsection.

Calculating the Coupling to Field Subsections in Non-Source Layers

With a source subsection A and a field subsection B, the System Matrixcoupling value is:

$\begin{matrix}{S_{AtoB} = {\int\limits_{V}{{E_{A} \cdot J_{B}}{dV}}}} & (14)\end{matrix}$

where E_(A) is the electric field due to the Source Subsection and J_(B)is the unit magnitude current distribution on the Field Subsection,i.e., the basis function. The integration is taken over the volume ofthe Field Subsection. If the Field Subsection is a surface currentinstead of a volume current, the integration is over the area of theField Subsection.

Applying (14) to (9), (10), (12), and (13), we have:

$\begin{matrix}{S_{AtoRFXU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{\int\limits_{V}{\left\{ {{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\}{J_{XYU} \cdot e_{i}}{dV}}}}}} & (15) \\{S_{AtoZU} = {\sum\limits_{TM}{\frac{V_{iU}^{(1)}k_{iz}^{(0)}N_{3}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{\int\limits_{V}{\left\{ {{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\} g_{3}{J_{ZU} \cdot u_{z}}{dV}}}}}} & (16) \\{S_{AtoXYL} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iT}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{\int\limits_{V}{\left\{ {{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{J_{XYL} \cdot e_{i}}{dV}}}}}} & (17) \\{S_{AtoZL} = {\sum\limits_{TM}{\frac{V_{iL}^{(1)}k_{iz}^{(2)}N_{3}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{\int\limits_{V}{\left\{ {{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{g_{3}\left( {x,y} \right)}{J_{ZL} \cdot u_{z}}{dV}}}}}} & (18)\end{matrix}$

where subscripts XY refer to a Field Subsection with transverse currentonly, Z refers to a Field Subsection with vertical current only, Urefers to the Upper Field Subsection, and L refers to a Lower FieldSubsection.

Coupling to Surface Rooftop Field Subsections

We evaluate the System Matrix coupling value by evaluating the integralportions of (15)-(18). For a RFX surface rooftop at the top of the UpperField Layer, above the Source Layer, at z=c₀, we work with (15):

$\begin{matrix}{S_{AtoRFXU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}r_{iT}^{(0)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}\underset{x,y}{\int\int}{\left\{ {{T\left( {x,x_{0},{\Delta\; x}} \right)}{R\left( {y,y_{0},{\Delta\; y}} \right)}u_{x}} \right\} \cdot {e_{i}\left( {x,y} \right)}}{dxdy}}}} & (19) \\{S_{AtoRFXU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}r_{iT}^{(0)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{T}\left( {\Delta\; x} \right)}{{F_{R}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{x}} \right\rbrack}}}} & (20)\end{matrix}$

See the (1) and (2) in the main paper for the definitions of the basisfunction, formed from triangle pulse functions, T(x, x₀, Δx) andrectangle pulse functions, R(y, y₀, Δy), and the Fourier coefficients,F_(T) (Δx) and F_(R) (Δy).

In an analogous manner, the System Matrix value for coupling to a RFYField Subsection above the Source Section is:

$\begin{matrix}{S_{AtoRFYU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}r_{iT}^{(0)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{{F_{T}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{y}} \right\rbrack}}}} & (21)\end{matrix}$

The differences from (20) are yellow highlighted in (21).

For coupling to a RFX or a RFY subsection at the bottom of the LowerField Layer, below the Source Layer, at z=c₃, we work with (17):

$\begin{matrix}{S_{AtoRFXL} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}r_{iB}^{(2)}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{T}\left( {\Delta\; x} \right)}{{F_{R}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{x}} \right\rbrack}}}} & (22) \\{S_{AtoRFYL} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}r_{iB}^{(2)}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{{F_{T}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{y}} \right\rbrack}}}} & (23)\end{matrix}$

Coupling to Volume Rooftop Field Subsections

For a VRFX Volume Rooftop at the top of the Upper Field Layer, above theSource Layer, from z=c₁ to c₀, we work with (15). The z-portion of therequired integral is:

$\begin{matrix}{\overset{c_{0}}{\int\limits_{z = c_{1}}}{\left\{ {{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\}{dz}}} & (24) \\{= {\frac{1}{k_{iz}^{(0)}}\left\lbrack {{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} - {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\rbrack}_{z = c_{1}}^{c_{0}}} & (25) \\{= {\frac{1}{k_{iz}^{(0)}}\left( {1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}h_{0}} \right)}}} \right)}} & (26)\end{matrix}$

This gives us:

$\begin{matrix}{S_{AtoVRFXU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}}{k_{iz}^{(0)}}\frac{1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}h_{0}} \right)}}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{T}\left( {\Delta\; x} \right)}{{F_{R}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{x}} \right\rbrack}}}} & (27)\end{matrix}$

In an analogous manner, the System Matrix value for coupling to a VRFYField Subsection above the Source Subsection is:

$\begin{matrix}{S_{AtoVRFYU} = {\sum\limits_{i}{\frac{V_{iU}^{(1)}}{k_{iz}^{(0)}}\frac{1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}h_{0}} \right)}}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{{F_{T}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{y}} \right\rbrack}}}} & (28)\end{matrix}$

For coupling to a VRFX subsection at the bottom of the Lower FieldLayer, below the Source Layer, from z=c₃ to c₂, we work with (17). Thez-portion of the required integral is:

$\begin{matrix}{\overset{c_{2}}{\int\limits_{z = c_{3}}}{\left\{ {{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{dz}}} & (29) \\{= {\frac{1}{k_{iz}^{(2)}}\left\lbrack {{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\rbrack}_{z = c_{3}}^{c_{2}}} & (30) \\{= {\frac{1}{k_{iz}^{(2)}}\left( {1 - {\cos\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}h_{2}} \right)}}} \right)}} & (31)\end{matrix}$

This gives us:

$\begin{matrix}{S_{AtoVRFXL} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}}{k_{iz}^{(2)}}\frac{1 - {\cos\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}h_{2}} \right)}}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{T}\left( {\Delta\; x} \right)}{{F_{R}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{x}} \right\rbrack}}}} & (32) \\{S_{AtoVRFYL} = {\sum\limits_{i}{\frac{V_{iL}^{(1)}}{k_{iz}^{(2)}}\frac{1 - {\cos\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}h_{2}} \right)}}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{{F_{T}\left( {\Delta\; y} \right)}\left\lbrack {{e_{i}\left( {x_{1},y_{1}} \right)} \cdot u_{y}} \right\rbrack}}}} & (33)\end{matrix}$

Coupling to Uniform Via Field Subsections

For a UV, Uniform Via, in the Upper Field Layer, above the Source Layer,from z=c₁ to c₀, we work with (16). The z-portion of the requiredintegral is:

$\begin{matrix}{\overset{c_{0}}{\int\limits_{z = c_{1}}}{\left\{ {{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)} + {r_{iT}^{(0)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\}{dz}}} & (34) \\{= {\frac{1}{k_{iz}^{(0)}}\left\lbrack {{- {\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}} - {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\rbrack}_{z = c_{1}}^{c_{0}}} & (35) \\{= {\frac{1}{k_{iz}^{(0)}}\left( {{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}\left( {{\cos\left( {k_{iz}^{(0)}h_{0}} \right)} - 1} \right)}} \right)}} & (36)\end{matrix}$

Completing the entire volume integral gives us:

$\begin{matrix}{S_{AtoUVU} = {\sum\limits_{TM}{V_{iU}^{(1)}N_{3}\frac{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}\left( {{\cos\left( {k_{iz}^{(0)}h_{0}} \right)} - 1} \right)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}}} & (37)\end{matrix}$

For coupling to a UV subsection at the bottom of the Lower Field Layer,below the Source Layer, from z=c₃ to c₂, we work with (18). Thez-portion of the required integral is:

$\begin{matrix}{\overset{c_{2}}{\int\limits_{z = c_{3}}}{\left\{ {{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} - {r_{iB}^{(2)}{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{dz}}} & (38) \\{= {\frac{1}{k_{iz}^{(2)}}\left\lbrack {{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)} - {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\rbrack}_{z = c_{3}}^{c_{2}}} & (39) \\{= {\frac{1}{k_{iz}^{(2)}}\left( {{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}\left( {{\cos\left( {k_{iz}^{(2)}h_{2}} \right)} - 1} \right)}} \right)}} & (40)\end{matrix}$

This gives us:

$\begin{matrix}{S_{AtoUVL} = {- {\sum\limits_{TM}{V_{iL}^{(1)}N_{3}\frac{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}\left( {{\cos\left( {k_{iz}^{(2)}h_{2}} \right)} - 1} \right)}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}}}} & (41)\end{matrix}$

Coupling to Tapered Via Field Subsections

For a TV, Tapered Via, at the top of the Upper Field Layer, above theSource Layer, from z=c₁ to c₀, we work with (16). The z-portion of thebasis function, J_(ZU), in (16) is (z−c₁)/h₀. ID-8b and ID-9b inAppendix 2 are helpful in evaluating the z-portion of the requiredintegral:

$\begin{matrix}{\frac{1}{h_{0}}{\overset{c_{0}}{\int\limits_{z = c_{1}}}{\left\{ {{\left( {z - c_{1}} \right){\cos\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}} - {{r_{iT}^{(0)}\left( {z - c_{1}} \right)}{\sin\left( {k_{iz}^{(0)}\left( {c_{0} - z} \right)} \right)}}} \right\}{dz}}}} & (42) \\{\mspace{79mu}{{{let}\text{:}\mspace{14mu} u} = {z - c_{1}}}} & (43) \\{= {{\frac{1}{h_{0}}{\overset{h_{0}}{\int\limits_{u = 0}}{\left\{ {{u\;{\cos\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}} - {r_{iT}^{(0)}u\;{\sin\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}}} \right\}{du}}}} = {\frac{1}{k_{iz}^{(0)}h_{0}}\left\lbrack {{{- u}\;{\sin\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}} + {\frac{1}{k_{iz}^{(0)}}{\cos\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}}} \right.}}} & (44) \\\left. \mspace{79mu}{{{- r_{iT}^{(0)}}u\;{\cos\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}} - {\frac{r_{iT}^{(0)}}{k_{iz}^{(0)}}{\sin\left( {k_{iz}^{(0)}\left( {h_{0} - u} \right)} \right)}}} \right\rbrack_{u = 0}^{h_{0}} & (45) \\{\mspace{79mu}{= {\frac{1}{k_{iz}^{(0)}h_{0}}\left( {{\frac{1}{k_{iz}^{(0)}}\left( {1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)}} \right)} - {r_{iT}^{(0)}h_{0}} + {\frac{r_{iT}^{(0)}}{k_{iz}^{(0)}}{\sin\left( {k_{iz}^{(0)}h_{0}} \right)}}} \right)}}} & (46) \\{\mspace{79mu}{= \frac{1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}\left( {{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} - {k_{iz}^{(0)}h_{0}}} \right)}}{k_{iz}^{{(0)}^{2}}h_{0}}}} & (47)\end{matrix}$

Completing the entire volume integral gives us:

$\begin{matrix}{S_{AtoTVU} = {\sum\limits_{TM}{\frac{V_{iU}^{(1)}N_{3}}{k_{iz}^{(2)}h_{2}}\frac{1 - {\cos\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}\left( {{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} - {k_{iz}^{(0)}h_{0}}} \right)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}}} & (48)\end{matrix}$

For a TV, Tapered Via, at the bottom of the Lower Field Layer, below theSource Layer, from z=c₃ to c₂, we work with (18). The z-portion of thebasis function, J_(ZU), in (16) is (z−c₃)/h₂. ID-8a and ID-9a inAppendix 2 are helpful in evaluating the z-portion of the requiredintegral:

$\begin{matrix}{\frac{1}{h_{2}}{\overset{c_{2}}{\int\limits_{z = c_{3}}}{\left\{ {{\left( {z - c_{3}} \right){\cos\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}} - {{r_{iB}^{(2)}\left( {z - c_{3}} \right)}{\sin\left( {k_{iz}^{(2)}\left( {z - c_{3}} \right)} \right)}}} \right\}{dz}}}} & (49) \\{\mspace{79mu}{{{let}\text{:}\mspace{14mu} u} = {z - c_{3}}}} & (50) \\{\mspace{79mu}{= {\overset{h_{2}}{\int\limits_{u = 0}}{\left\{ {{u\;{\cos\left( {k_{iz}^{(2)}u} \right)}} - {r_{iB}^{(2)}u\;{\sin\left( {k_{iz}^{(2)}u} \right)}}} \right\}{du}}}}} & (51) \\{\mspace{79mu}\begin{matrix}{= \left( {\frac{1}{k_{iz}^{(2)}h_{2}}\left\lbrack {{u\;{\sin\left( {k_{iz}^{(2)}u} \right)}} + {\frac{1}{k_{iz}^{(0)}}\cos\left( {k_{iz}^{(2)}u} \right)} +} \right.} \right.} \\\left. {{r_{iB}^{(2)}u\;{\cos\left( {k_{iz}^{(2)}u} \right)}} - {\frac{r_{iB}^{(2)}}{k_{iz}^{(0)}}{\sin\left( {k_{iz}^{(2)}u} \right)}}} \right\rbrack_{u = 0}^{h_{2}}\end{matrix}} & (52) \\\left. \mspace{79mu}{{- \frac{r_{iB}^{(2)}}{k_{iz}^{(0)}}}{\sin\left( {k_{iz}^{(2)}h_{2}} \right)}} \right) & (53) \\{\mspace{79mu}{= \frac{\begin{matrix}{{\left( {1 + {r_{iB}^{(2)}k_{iz}^{(2)}h_{2}}} \right){\cos\left( {k_{iz}^{(2)}h_{2}} \right)}} -} \\{1 + {\left( {{k_{iz}^{(2)}h_{2}} - r_{iB}^{(2)}} \right){\sin\left( {k_{iz}^{(2)}h_{2}} \right)}}}\end{matrix}}{k_{iz}^{{(2)}^{2}}h_{2}}}} & (54)\end{matrix}$

Completing the entire volume integral gives us:

$\begin{matrix}{S_{AtoTVL} = {- {\sum\limits_{TM}{\frac{V_{iL}^{(1)}N_{3}}{k_{iz}^{(2)}h_{2}}\frac{\begin{matrix}{{\left( {1 + {r_{iB}^{(2)}k_{iz}^{(2)}h_{2}}} \right){\cos\left( {k_{iz}^{(2)}h_{2}} \right)}} -} \\{1 + {\left( {{k_{iz}^{(2)}h_{2}} - r_{iB}^{(2)}} \right){\sin\left( {k_{iz}^{(2)}h_{2}} \right)}}}\end{matrix}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iB}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}{g_{3}\left( {x_{1},y_{1}} \right)}}}}} & (55)\end{matrix}$

Coupling from Volume Rooftop Source Subsections

To determine coupling from the Source Subsection, we must determine theV_(iU) ⁽¹⁾ for coupling to Field Subsections above the Source Layer, andthe V_(iL) ⁽¹⁾ for coupling to Field Subsections below the Source Layer.We provide these results in this section and the next two sections.

The transverse fields from an x-directed volume rooftop, VRFX, and ay-directed volume rooftop, VRFY, are given in (60) in the VolumeRooftopsdocument, repeated here with Layer indices added. The C_(i) come from(27) and (29) in the VolumeRooftops document:

$\begin{matrix}{E_{tVRF} = {\sum\limits_{i}{\frac{C_{i}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\left( {1 - \frac{\begin{matrix}{{\sin\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)} + {r_{iT}^{(1)}\cos}} \\{{\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)\sin\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)} +} \\{r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}} \right)e_{i}}}} & (56)\end{matrix}$

where

C _(i)(x ₀ ,y ₀)=F _(T)(Δx)F _(R)(Δy)[e _(i)(x ₀ ,y ₀)·u _(x)]  (57) forVRFX, and

C _(i)(x ₀ ,y ₀)=F _(R)(Δx)F _(T)(Δy)[e _(i)(x ₀ ,y ₀)·u _(y)]  (58) forVRFY

By applying (1) to (56), we evaluate the transverse fields on the upper,z=c₁, surface of the Source Layer:

$\begin{matrix}{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\left( {1 - \frac{r_{iT}^{(1)} + {\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}} \right)}} & (59) \\{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\frac{\begin{matrix}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + \left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right)} \\{{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - r_{iT}^{(1)} - {\sin\left( {k_{iz}^{(1)}h_{1}} \right)} - {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}}} & (60) \\{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\frac{r_{iT}^{(1)}\left( {{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1 - {r_{iB}^{(1)}{\sin\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}}} & (61)\end{matrix}$

Either (59) or (61) can be used, depending on which is most convenient.

Next we apply (11) to (56), we evaluate the transverse fields on thelower, z=c₂, surface of the Source Layer:

$\begin{matrix}{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\left( {1 - \frac{{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iT}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}} + r_{iB}^{(1)}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}} \right)}} & (62) \\{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\left( \frac{\begin{matrix}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + \left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right)} \\{{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - {\sin\left( {k_{iz}^{(1)}h_{1}} \right)} - {r_{iT}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}} - r_{iB}^{(1)}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (63) \\{V_{iVRFU}^{(1)} = {\frac{C_{is}\left( {x_{0},y_{0}} \right)}{Y_{i}^{(1)}k_{iz}^{(1)}}\left( \frac{r_{iB}^{(1)}\left( {{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1 - {r_{iT}^{(1)}{\sin\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (64)\end{matrix}$

Either (62) or (64) can be used, depending on which is most convenient.

Coupling from Uniform Via Source Subsections

The transverse fields from a Uniform Via, UV, are given in (141a) and(124) in the ViasNew document, and repeated here with Layer indicesadded:

$\begin{matrix}{E_{tUV} = {\sum\limits_{TM}{\frac{T_{it}}{k_{iz}^{(1)}}\left\{ {{- {r_{iB}^{(1)}\left( {{\sin\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)} + {r_{iT}^{(1)}{\cos\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)}}} \right)}} + {r_{iT}^{(1)}\left( {{\sin\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)}}} \right)}} \right\} e_{i}}}} & (65) \\{\mspace{79mu}{T_{it} = {\frac{k_{iz}^{(1)}}{Y_{i}^{(1)}}\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{{\left( {1 - {r_{iT}^{(1)}r_{iB}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}}}} & (66)\end{matrix}$

By applying (1) to (65), we evaluate the transverse fields on the upper,z=c₁, surface of the Source Layer:

$\begin{matrix}{V_{iUVU}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}}\left( \frac{r_{iT}^{(1)}\left( {{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iB}^{(1)}\left( {{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1} \right)}} \right)}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (67)\end{matrix}$

By applying (11) to (65), we evaluate the transverse fields on thelower, z=c₂, surface of the Source Layer:

$\begin{matrix}{V_{iUVU}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}}\left( \frac{r_{iB}^{(1)}\left( {{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iT}^{(1)}\left( {{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1} \right)}} \right)}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (68)\end{matrix}$

Coupling from Tapered Via Source Subsections

The transverse fields from a Tapered Via, TV, are given in (190) inAppendix 1, and repeated here with Layer indices added:

$\begin{matrix}{E_{tTV} = {\sum\limits_{TM}{\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( \frac{\begin{matrix}{1 - {\sin\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)} + {r_{iT}^{(1)}\cos\left( {k_{iz}^{(1)}\left( {c_{1} - z} \right)} \right)} +} \\{\left( {1 - {r_{iT}^{(1)}k_{iz}^{(1)}h_{1}}} \right)\left( {{\sin\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}\left( {z - c_{2}} \right)} \right)}}} \right)}\end{matrix}}{{\left( {1 - {r_{iT}^{(1)}r_{iB}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)e_{i}}}} & (68)\end{matrix}$

By applying (1) to (68), we evaluate the transverse fields on the upper,z=c₁, surface of the Source Layer:

$\begin{matrix}{V_{iTVU}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( {1 - \frac{r_{iT}^{(1)} + {\left( {1 - {r_{iT}^{(1)}k_{iz}^{(1)}h_{1}}} \right)\left( {{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right.}}}} \right)}} & \left( {69a} \right) \\{V_{iTVU}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( \frac{\begin{matrix}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right)\cos\left( {k_{iz}^{(1)}h_{1}} \right)} -} \\{r_{iT}^{(1)} - {\left( {1 - {r_{iT}^{(1)}k_{iz}^{(1)}h_{1}}} \right)\left( {{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & \left( {69b} \right) \\{V_{iTVU}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( {r_{iT}^{(1)}\frac{\begin{matrix}{{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1 - {r_{iB}^{(1)}\sin\left( {k_{iz}^{(1)}h_{1}} \right)} +} \\{k_{iz}^{(1)}{h_{1}\left( {{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iB}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}} \right)}} & \left( {69c} \right)\end{matrix}$

Use either (69a) or (69c), whichever is most convenient.

By applying (11) to (68), we evaluate the transverse fields on thelower, z=c₂, surface of the Source Layer:

$\begin{matrix}{V_{iTVL}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( {1 - \frac{{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} + {r_{iT}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}} + {r_{iB}^{(1)}\left( {1 - {r_{iT}^{(1)}\left( {k_{iz}^{(1)}h_{1}} \right)}} \right.}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}}} \right)}} & (70) \\{V_{iTVL}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( \frac{\begin{matrix}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right)\cos\left( {k_{iz}^{(1)}h_{1}} \right)} -} \\{{\sin\left( {k_{iz}^{(1)}h_{1}} \right)} - {r_{iT}^{(1)}{\cos\left( {k_{iz}^{(1)}h_{1}} \right)}} - {r_{iB}^{(1)}\left( {1 - {r_{iT}^{(1)}k_{iz}^{(1)}h_{1}}} \right)}}\end{matrix}}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (71) \\{V_{iTVL}^{(1)} = {\frac{{F_{R}\left( {\Delta\; x} \right)}{F_{R}\left( {\Delta\; y} \right)}N_{3}{g_{3}\left( {x_{0},y_{0}} \right)}}{Y_{i}^{(1)}k_{iz}^{(1)}h_{1}}\left( \frac{r_{iB}^{(1)}\left( {{\cos\left( {k_{iz}^{(1)}h_{1}} \right)} - 1 - {r_{iT}^{(1)}{\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {r_{iT}^{(1)}k_{iz}^{(1)}h_{1}}} \right)}{{\left( {1 - {r_{iB}^{(1)}r_{iT}^{(1)}}} \right){\sin\left( {k_{iz}^{(1)}h_{1}} \right)}} + {\left( {r_{iT}^{(1)} + r_{iB}^{(1)}} \right){\cos\left( {k_{iz}^{(1)}h_{1}} \right)}}} \right)}} & (72)\end{matrix}$

Either (70) or (72) can be used, depending on which is most convenient.

Calculating Coupling Across Multiple Layers

To calculate coupling to a Field Subsection that is in a layerimmediately above or below the Source Subsection, select the desired‘Coupling From’ and ‘Coupling To’ subsection type. Then select theappropriate ‘Coupling To’ section and ‘Coupling From’ section. Then, inthe ‘Coupling From’ section, evaluate V_(iU) ⁽¹⁾ (for the layer above,i.e., smaller index), or V_(iL) ⁽¹⁾ (for the layer below, or largerindex). Use the selected expression in the appropriate equation in the“Coupling To” sections.

The V_(iU) ⁽¹⁾ correspond to the tangential electric fields on the topsurface of the Source Layer. If the Field Layer is higher still, use (8)to determine the source fields in Field Layer, then treat those fieldsas the new source fields and determine the V_(i) for the next layer up.Evaluate the tangential fields at the top of that layer and repeat untilthe desired Field Layer is reached. This recursive procedure issummarized next.

-   -   1) Determine the desired V_(iU) ⁽¹⁾ for the desired Source        Subsection.    -   2) If the Field Subsection is in the layer immediately above,        use that V_(iU) ⁽¹⁾ in the appropriate Coupling To equation.    -   3) Otherwise, transform the V_(iU) ⁽¹⁾ to the next layer up        (i.e., smaller index, here it is 0) using:

$\begin{matrix}{V_{iU}^{(0)} = \frac{V_{iU}^{(1)}r_{iT}^{(0)}}{{\sin\left( {k_{iz}^{(0)}h_{0}} \right)} + {r_{iT}^{(0)}{\cos\left( {k_{iz}^{(0)}h_{0}} \right)}}}} & (73)\end{matrix}$

-   -   4) Use the V_(iU) ⁽⁰⁾ calculated above as the new V_(iU) ⁽¹⁾ and        proceed to step 2.

If the Field Subsection Layer is below the Source Subsection Layer,proceed in an analogous manner, using

$\begin{matrix}{V_{iL}^{(2)} = \frac{V_{iL}^{(1)}r_{iT}^{(2)}}{{\sin\left( {k_{iz}^{(2)}h_{2}} \right)} + {r_{iT}^{(2)}{\cos\left( {k_{iz}^{(2)}h_{2}} \right)}}}} & (74)\end{matrix}$

Appendix 4

Appendix 4: Volume Current Ohmic Loss

Current on each subsection is represented in terms of a unit amplitudecurrent distribution known as a basis function. For the following,subsection A is the source subsection with basis function J_(A), andsubsection B is the field subsection with basis function J_(B). Themoment matrix element for coupling from subsection A to subsection B is

$\begin{matrix}{S_{AtoB} = {\int\limits_{V}{{E_{A} \cdot J_{B}}{dV}}}} & (1)\end{matrix}$

where E_(A) is the electric field generated by the source subsection.Weighting the electric field by the field basis function, J_(B), givesus a Galerkin implementation of the method of moments, which is used inSonnet. Due to reciprocity, the Galerkin technique yields a symmetricmatrix. To calculate Ohmic loss in a conductor, we evaluate only theelectric field generated by the Ohmic loss. Thus, we assume the Asubsection is a conductor with a bulk conductivity of a that obeys thefollowing relationship:

$\begin{matrix}{E_{A} = \frac{J_{A}}{\sigma}} & (2)\end{matrix}$

In general, a may be complex, for example, to model kinetic inductancein superconductors. Substituting (2) into (1):

$\begin{matrix}{S_{AtoB} = {\int\limits_{V}{\frac{J_{B}{‘J_{A}}}{\sigma}dV}}} & (3)\end{matrix}$

The above integral has a non-zero value only when the A and Bsubsections are one and the same or when they are physicallyoverlapping. Even if overlapping, the integral is still zero if thecurrent on each subsection is orthogonal to the current on the other. Ifboth subsections support only surface current, rather than volumecurrent, then the integral reduces to a surface integral and the bulkconductivity, a, becomes a surface conductivity.

Care should be taken to make sure a surface current subsection does notrepresent exactly the same volume of conductor that a volume currentsubsection represents. For example, if a surface subsection is placeddirectly on top of a volume subsection with both representing currentflowing in the same direction, but the intent is for the surfacesubsection to represent a volume of conductor above the volumesubsection, there should be no problem. However, if the intent of thesurface subsection is to represent the same volume of conductor as thevolume subsection immediately below it, the total effective conductivitywould then be double the desired conductivity as the two subsectionswould be connected in parallel.

If a subsection supports only surface current and it overlaps a volumecurrent subsection, the Ohmic coupling in the overlapping region shouldbe treated exactly the same for both subsections within the region ofoverlap. For example, the overlapping region could be treated as avolume current for both with bulk conductivity a. Alternatively, bothcould be treated as a surface current subsection with a specifiedsurface conductivity. The important thing is that reciprocity hold. Inother words, the Ohmic coupling from A to B should be exactly equal tothe Ohmic coupling from B to A. This keeps the system matrix symmetric.

Rooftop Ohmic Self-Coupling

The (symmetric) volume rooftop basis function for x-directed current,with its base centered at (x₀, y₀, z₀)=(0 0,0) is:

$\begin{matrix}{{{J_{VRFX}\left( {0,0,0} \right)} = {1 - \frac{x}{2\Delta\; x}}},{0 \leq x \leq {2\Delta\; x}},{{{- \Delta}\; y} \leq y \leq {\Delta\; y}},{0 \leq z \leq h}} & \left( {4a} \right) \\{\mspace{79mu}{{= {1 + \frac{x}{2\Delta\; x}}},{{{- 2}\Delta\; x} \leq x \leq 0},{{{- \Delta}\; y} \leq y \leq {\Delta\; y}},{0 \leq z \leq h}}} & \left( {4b} \right) \\{\mspace{79mu}{{= 0},{otherwise}}} & \left( {4c} \right)\end{matrix}$

Since (4) is symmetric, we need only integrate (4a) and double theresult. Thus, the Ohmic self-coupling, using (3), yields:

$\begin{matrix}{S_{VRFXtoSelf} = {\frac{2}{\sigma}{\int_{x = 0}^{2\Delta\; x}{\int_{y = {{- \Delta}\; y}}^{\Delta\; y}{\int_{z = 0}^{h}{\left( {1 - \frac{x}{2\Delta\; x}} \right)^{2}{dxdydz}}}}}}} & (5) \\{= {{\frac{4\Delta\;{yh}}{\sigma}{\int_{x = 0}^{2\Delta\; x}1}} - \frac{x}{\Delta\; x} + {\frac{x^{2}}{4\Delta\; x^{2}}{dx}}}} & (6) \\{= {\frac{4\Delta\;{yh}}{\sigma}\left\lbrack {x - \frac{x^{2}}{2\Delta\; x} + \frac{x^{3}}{12\Delta\; x^{2}}} \right\rbrack}_{x = 0}^{x = {2\Delta\; x}}} & (7) \\{= {\frac{4\Delta\;{yh}}{\sigma}\left( {{2\Delta\; x} - {2\Delta\; x} + {2\Delta\;{x/3}}} \right)}} & (8) \\{= \frac{8\Delta\; x\;\Delta\;{yh}}{3\sigma}} & (9)\end{matrix}$

The x-directed rooftop is composed of two adjacent cells. The left cellis called ‘ramp-up’ and the right cell is called ‘ramp-down’. They-directed rooftop has a similar structure, only along the y-axis Eachof the two cells have base dimensions of 2Δx by 2Δy. Thus, we can put(9) in terms of cell volume. In addition, the self-coupling for ay-directed rooftop that has the same cell volume is identical:

$\begin{matrix}{V_{Cell} = {{\left( {2\Delta\; x} \right)\left( {2\Delta\; y} \right)h} = {4\Delta\; x\;\Delta\;{yh}}}} & (10) \\{S_{VRFXtoSelf} = {S_{VRFYtoSelf} = \frac{2V_{Cell}}{3\sigma}}} & (11)\end{matrix}$

For surface rooftop basis function self-coupling, we do not perform theintegration over z in (5). Thus, we can modify (11) to use the area of acell and a surface conductance:

$\begin{matrix}{A_{Cell} = {{\left( {2\Delta\; x} \right)\left( {2\Delta\; y} \right)h} = {4\Delta\; x\;\Delta\; y}}} & (12) \\{S_{RFXtoSelf} = {S_{RFYtoSelf} = \frac{2A_{Cell}}{3\sigma_{S}}}} & (13)\end{matrix}$

Rooftop Ohmic Adjacent-Coupling

Two rooftop basis functions are considered adjacent if one eaves of eachroof-top touches the peak of the other. Two adjacent same-directedrooftops overlap over half of each rooftop, i.e., over the area of onecell. For example a rooftop centered at (0,0,0) and a second one at(2Δx, 0,0) overlap over the range of 0≤x≤2Δx. It is only thisoverlapping region where Ohmic coupling can take place. The currentdistribution for the first rooftop in this example is given in (4). The(symmetric) volume rooftop basis function for x-directed current,centered at (x₀, y₀, z₀)=(2Δx, 0,0) is:

$\begin{matrix}{{{J_{VRFX}\left( {{2\Delta\; x},{0,0}} \right)} = \frac{x}{2\Delta\; x}},{0 \leq x \leq {2\Delta\; x}},{{{- \Delta}\; y} \leq y \leq {\Delta\; y}},{0 \leq z \leq h}} & \left( {14a} \right) \\{{= {2 - \frac{x}{2\;\Delta\; x}}},{{2\Delta\; x} \leq x \leq {4\Delta\; x}},{{{- \Delta}\; y} \leq y \leq {\Delta\; y}},{0 \leq z \leq h}} & \left( {14b} \right) \\{{= 0},{otherwise}} & \left( {14c} \right)\end{matrix}$

Since this rooftop and the rooftop of (4) overlap only over the range0≤x≤2Δx, that is the range over which we integrate. Substituting (4a)and (14a) into (3), we evaluating the Ohmic adjacent-coupling as:

$\begin{matrix}{S_{VRFXtoSelf} = {\frac{1}{\sigma}{\int_{x = 0}^{2\Delta\; x}{\int_{y = {{- \Delta}\; y}}^{\Delta\; y}{\int_{z = 0}^{h}{\left( {1 - \frac{x}{2\Delta\; x}} \right)\frac{x}{2\Delta\; x}{dxdydz}}}}}}} & (15) \\{= {{\frac{2\Delta\;{yh}}{\sigma}{\int_{x = 0}^{2\Delta\; x}\frac{x}{2\Delta\; x}}} - {\frac{x^{2}}{4\Delta\; x^{2}}{dx}}}} & (16) \\{= {\frac{2\Delta\;{yh}}{\sigma}\left\lbrack {\frac{x^{2}}{4\Delta\; x} + \frac{x^{3}}{12\Delta\; x^{2}}} \right\rbrack}_{x = 0}^{x = {2\Delta\; x}}} & (17) \\{= {\frac{2\Delta\;{yh}}{\sigma}\left( {\frac{x^{2}}{4\Delta\; x} + \frac{x^{3}}{12\Delta\; x^{2}}} \right)}} & (18) \\{= \frac{2\Delta\; x\;\Delta\;{yh}}{3\sigma}} & (19)\end{matrix}$

Using (10), we see that the adjacent subsection Ohmic coupling is:

$\begin{matrix}{S_{VRFXtoAdj} = {S_{VRFYtoAdj} = \frac{V_{Cell}}{6\sigma}}} & (20)\end{matrix}$

For surface rooftop basis function self-coupling, we do not perform theintegration over z in (15). Thus, we can modify (20) to use the area ofa cell and a surface conductance:

$\begin{matrix}{S_{RFXtoAdj} = {S_{RFYtoAdj} = \frac{A_{Cell}}{6\sigma_{S}}}} & (21)\end{matrix}$

Uniform Via Ohmic Self-Coupling

Since the uniform via has uniform z-directed over its entire volume (onecell), the integral of (3) results in:

$\begin{matrix}{S_{UVIAtoSelf} = \frac{V_{Cell}}{\sigma_{S}}} & (22)\end{matrix}$

Uniform Via to Tapered Via Ohmic Coupling

The uniform via has uniform z-directed current over its entire volume.The tapered via, with its lower end centered at (0,0,0) has thefollowing current distribution:

$\begin{matrix}{{{J_{TVIA}\left( {0,0,0} \right)} = \frac{z}{h}},{{{- \Delta}\; x} \leq x \leq {\Delta\; x}},{{{- \Delta}\; y} \leq y \leq {\Delta\; y}},{0 \leq z \leq h}} & \left( {23a} \right) \\{{= 0},{otherwise}} & \left( {23b} \right)\end{matrix}$

The integral of (3) results in:

$\begin{matrix}{S_{UVIAtoTVIA} = {\frac{1}{\sigma}{\int_{x = {{- \Delta}x}}^{\Delta x}{\int_{y = {{- \Delta}y}}^{\Delta y}{\int_{z = 0}^{h}{\frac{z}{h}{dxdydz}}}}}}} & (24) \\{= {\frac{4\Delta x\Delta y}{\sigma}{\int_{z = 0}^{h}{\frac{z}{h}{dx}}}}} & (25) \\{= {\frac{4\Delta x\Delta y}{\sigma}\left\lbrack \frac{z^{2}}{2h} \right\rbrack}_{x = 0}^{x = h}} & (26) \\{= {\frac{4\Delta x\Delta y}{\sigma}\left( \frac{h^{2}}{2h} \right)}} & (27) \\{= \frac{2\Delta\; x\;\Delta\;{yh}}{\sigma}} & (28)\end{matrix}$

Putting this result in terms of the cell volume, (10):

$\begin{matrix}{S_{UVIAtoTVIA} = \frac{V_{Cell}}{2\sigma_{S}}} & (29)\end{matrix}$

Keep in mind that this result is valid only for UVIAs and TVIAs thatshare the same cell. All other UVIA to TVIA Ohmic-couplings are zero.

Tapered Via Ohmic Self-Coupling

Applying (23a) for both current distributions, the integral of (3)results in:

$\begin{matrix}{S_{TVIAtoSelf} = {\frac{1}{\sigma}{\int_{x = {{- \Delta}x}}^{\Delta x}{\int_{y = {{- \Delta}y}}^{\Delta y}{\int_{z = 0}^{h}{\frac{z^{2}}{h^{2}}{dxdydz}}}}}}} & (30) \\{= {\frac{4\Delta x\Delta y}{\sigma}{\int_{z = 0}^{h}{\frac{z^{2}}{h^{2}}{dx}}}}} & (31) \\{= {\frac{4\Delta x\Delta y}{\sigma}\left\lbrack \frac{z^{3}}{3h^{2}} \right\rbrack}_{x = 0}^{x = h}} & (32) \\{= {\frac{4\Delta x\Delta y}{\sigma}\left( \frac{h^{3}}{3h^{2}} \right)}} & (33) \\{= \frac{4\Delta\; x\;\Delta\;{yh}}{3\sigma}} & (34)\end{matrix}$

Putting this result in terms of the cell volume, (10):

$\begin{matrix}{S_{TVIAtoSelf} = \frac{V_{Cell}}{3\sigma_{S}}} & (35)\end{matrix}$

All Other Ohmic Couplings

All other Ohmic couplings are zero because all other possible pairs ofsubsections do not have any component of current (J_(x), J_(y), orJ_(z)) that are in common with each other. In other words, the dotproduct of (3) is zero for all other cases.

What is claimed is:
 1. A computer-implemented method of simulating avertically-oriented current distribution of current flowing through aplurality of layers of a three-dimensional structure embedded in ashielded multi-layered dielectric, comprising the steps of: dividing thestructure into a plurality of subsections, wherein at least one portionof the structure having vertically-oriented current is divided into atleast one rectangular prism subsection having a first surface disposedperpendicular to a z-axis and a second surface disposed perpendicular tothe z-axis; independently assigning a current in each of thesubsections, wherein z-directed current in the at least one rectangularprism subsection linearly changes from a first value at the firstsurface of the rectangular prism subsection to a second value at thesecond surface of the rectangular prism subsection; wherein the firstvalue is different from the second value; calculating an induced voltagein each of the plurality of subsections resulting from eachindependently modeled current, the induced voltage in each of theplurality of subsections corresponding to a transfer impedance ortransfer admittance; and calculating a current distribution in one ormore conductors according to the transfer impedance or transferadmittance of each subsection and an assumed voltage across eachsubsection.
 2. The computer-implemented method of claim 1, wherein theassumed voltage is zero except where a voltage source is coupled.
 3. Thecomputer-implemented method of claim 1, wherein the assumed voltage isproportional to the current in the subsection.
 4. Thecomputer-implemented method of claim 1, wherein the first value ishigher than the second value.
 5. The computer-implemented method ofclaim 1, wherein the first value is lower than the second value.
 6. Thecomputer-implemented method of claim 1, wherein z-directed current in asecond rectangular prism subsection is constant throughout.
 7. Thecomputer-implemented method of claim 1, wherein both the firstpredetermined value and second predetermined value are nonzero.
 8. Thecomputer-implemented method of claim 1, wherein at least one portion ofthe structure carrying horizontal current is divided into volume rooftopsubsections.
 9. The computer-implemented method of claim 1, wherein atleast one portion of the structure carrying horizontal current isdivided into rooftop subsections.
 10. The computer-implemented method ofclaim 1, wherein the current distribution is calculated via a matrixinversion.
 11. A computer program product stored on a non-transitorycomputer-readable medium which includes a set of non-transitorycomputer-readable instructions for simulating a vertically-orientedcurrent distribution of current flowing through a plurality of layers ofa three-dimensional structure embedded in a shielded multi-layereddielectric, the instructions comprising the steps of: dividing thestructure into a plurality of subsections, wherein at least one portionof the structure having vertically-oriented current is divided into atleast one rectangular prism subsection having a first surface disposedperpendicular to a z-axis and a second surface disposed perpendicular tothe z-axis; independently assigning a current in each of thesubsections, wherein z-directed current in the at least one rectangularprism subsection linearly changes from a first value at the firstsurface of the rectangular prism subsection to a second value at thesecond surface of the rectangular prism subsection; wherein the firstvalue is different from the second value; calculating an induced voltagein each of the plurality of subsections resulting from eachindependently modeled current, the induced voltage in each of theplurality of subsections corresponding to a transfer impedance ortransfer admittance; and calculating a current distribution in one ormore conductors according to the transfer impedance or transferadmittance of each subsection and an assumed voltage across eachsubsection.
 12. The computer program product of claim 11, wherein theassumed voltage is zero except where a voltage source is coupled. 13.The computer program product of claim 11, wherein the assumed voltage isproportional to the current in the subsection.
 14. The computer programproduct of claim 11, wherein the first value is higher than the secondvalue.
 15. The computer program product of claim 11, wherein the firstvalue is lower than the second value.
 16. The computer program productof claim 11, wherein z-directed current in a second rectangular prismsubsection is constant throughout.
 17. The computer program product ofclaim 11, wherein both the first predetermined value and secondpredetermined value are nonzero.
 18. The computer program product ofclaim 11, wherein at least one portion of the structure carryinghorizontal current is divided into volume rooftop subsections.
 19. Thecomputer program product of claim 11, wherein at least one portion ofthe structure carrying horizontal current is divided into rooftopsubsections.
 20. The computer program product of claim 11, wherein thecurrent distribution is calculated via a matrix inversion.